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### Course: Trigonometry > Unit 4

Lesson 6: Challenging trigonometry problems- Trig challenge problem: area of a triangle
- Trig challenge problem: area of a hexagon
- Trig challenge problem: cosine of angle-sum
- Trig challenge problem: arithmetic progression
- Trig challenge problem: maximum value
- Trig challenge problem: multiple constraints
- Trig challenge problem: system of equations

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# Trig challenge problem: multiple constraints

Sal solves a very complicated algebraic trig problem that appeared as problem 47 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.

## Want to join the conversation?

- Dear Sir,

About the middle of your solution, I think, we should have considered: 2(thita) = (pi/2)- 2(thita)[NOT: 4(thita)] + 2n(pi)(5 votes)- That would be assuming a 45deg angle: if we change the pi/2 from radians to degrees, we get 2theta = 90 - 2theta OR x = 90-x.

Sal did not randomly put in 4 where he should have put 2 - he used the given equation "sin 2theta = cos 4theta" to get a sin = sin equation. This is btwn8:00and8:30in the video.(6 votes)

- Is IIT for India only.

How hard is the IIT JEE?

Can I get admission into IIT Mumbai with 70 percent in IIT?(4 votes)- Yeah, You can get into IITB easily if you score 70% (252 off 360)

Jee advanced 2014 mark vs rank: http://www.embibe.com/100marks/News/jee-advanced-marks-vs-ranks/

2014 cutoff for different IIT courses=>http://cutoffs.aglasem.com/jee-advanced-cut-off-iits/(8 votes)

- I agree that sin(x) = sin(x+2pi), but sin(x) = sin(pi-x) also, however those aren't taken into account in this video....???? Am I missing something?(4 votes)
- The sin(x) DOES always = sin( pi-x ). This is simple trig identity, easy to proof. It holds for the given example of x=5pi/4, since sin(5pi/4)=-sqrt(2)/2=sin(-pi/4)=sin(pi-5pi/4). This identity indeed should be taken into account. And only by accident the solution found by Sal is correct. This is because the identity sin(x)=sin( pi-x ) does not provide any more solutions it this particular case which is considered in the film, but in other problems it can provide more solutions, and always should be taken into account. Consider for example the equation sin(x)=sin(2x). In this case Sal's method will not give all solutions to the equation, but only part of them, the rest can be found using the identity sin(x)=sin(pi-x).(4 votes)

- Is this trig trig or is this trig calculus? I'm taking trig should i worry if i don't get this?(3 votes)
- This is trig but I wouldn't worry about not getting it, these are really difficult problems.(4 votes)

- i just don't understand y we have to add npie(3 votes)
- We need to find all values of theta, you don't (yet) know how many are there.

if you just solve for 3/4 = tan(theta), you can can find one of them, but not all. So the question is what are all the possible values of theta that will give tan(of each theta) to be 3/4. So you use the identity that tan(theta) = tan(theta + n*pi)(3 votes)

- What level math is this?? is it precalculus? Because it's way too difficult for me.(2 votes)
- @4:39your saying cot 5 theta + npi = tan (pi/2 - 5 theta + n pi) which is arbitrary. I found another who agrees....

"Dear Sir,

About the middle of your solution, I think, we should have considered: 2(thita) = (pi/2)- 2(thita)[NOT: 4(thita)] + 2n(pi)"

2 Votes • Flag 8 months ago by a_ashkenazi(1 vote)- Because cot = cos/sin and tan = sin/cos, if you do cot of one angle (of a right triangle) then it is the same as the tan of the other angle. Another way of writing this is:

cot(x) = tan (90-x) for degrees or cot(x) = tan (pi/2 - x) for radians. Then substitute 5theta for x.

We add the n*pi because at every cycle +pi tan (and cot) would be the same. If the range is limited to -pi/2 to pi/2, why would we add the n*pi? If we go back to graphing trig functions and graph sin(5theta) for example, we'll see 5 cycles of sin btwn 0 and 2pi instead of one. So tan(theta) = cot(5theta) and sin(2theta) = cos(4theta) means we'll have to consider the wider range until we get our final answers. If you were to graph them each (sin, cos, tan, cot, continuously both directions) then you'd see the 3 points in the given range at the end.(3 votes)

- So, I know that in chemistry, the units can usually be treated like variables. Is this the same for trig functions? At5:17, it looks almost like Sal divided both sides by the tangent. Is that what he did, or was he just concluding that (theta) and (pi over 2 + 5 theta + n theta) must be equal to each other because their tangents were equal to each other?(2 votes)
- You can cancel out the effect of a tangent by applying the arctangent function to both sides. Arctan(tan(x)) = x. By using the inverse function on both sides, you can cancel out both of the tangents and just leave the inputs. It's like exponents and logarithms, or squares and square roots.(1 vote)

- why is that only 3 values of theta? instead of 9 since 6 for the tan theta = cot 5theta and 3 for sin 2theta = cos 4theta. I think what we really want to determine is that the number of common values of 2 equations subject to restriction.(1 vote)
- 𝜃 ∈ {−5𝜋∕12, −3𝜋∕12, −𝜋∕12, 𝜋∕12, 3𝜋∕12, 5𝜋∕12}

These six values are the only values in the interval (−𝜋∕2, 𝜋∕2)

that also solve the equation tan 𝜃 = cot 5𝜃

– – –

Similarly, we have three values,

𝜃 ∈ {−3𝜋∕12, 𝜋∕12, 5𝜋∕12},

that are in the interval (−𝜋∕2, 𝜋∕2)

and solve the equation sin 2𝜃 = cos 4𝜃

– – –

Now, the second constraint is that both equations must be true.

𝜃 ∈ {−5𝜋∕12, −𝜋∕12, 3𝜋∕12} don't meet this constraint because they only solve one of the equations.

So, that leaves us with the three values:

𝜃 ∈ {−3𝜋∕12, 𝜋∕12, 5𝜋∕12}

– – –

There is also a third constraint that says that none of these values can be a multiple of 𝜋∕5, but since none of them are we're still left with three values.(3 votes)

- The first question is:

if sin(2Θ) = sin( pi/2 - 4Θ +2pi*n), but sin(pi/2 -4Θ +2pi*n) = sin(pi/2 + 4Θ +2pi*n), because sin(pi/2 - x) = sin(pi/2 + x), therefore it shouldn't be taken to consideration?

The second:

Therefore the answer are the three angles in radians?(2 votes)

## Video transcript

The number of values of theta
in the interval from negative pi over 2 to pi over
2-- and it's not including those because we have
curly parentheses around it --such that theta does
not equal n pi over 5. So it's not a
multiple of pi over 5, for n equals 0 plus or
minus 1, plus or minus 2. And tangent of theta is equal
to cotangent of 5 theta, as well as sine of 2 theta is
equal to cosine of 4 theta is-- We have to find the
number of values of theta that satisfy all
of these things. So just from the
get-go, it seems like we're going to
have to solve for thetas and convert between sine and
cosine which is essentially what you're doing between
tangent and cotangent. We'll do that in a little bit. And those of you who
know, if you're actually taking an exam under
time pressure, I don't recommend you reproving
your trig identities from first principles. But for education purposes,
I always like to do it. Let me just draw a right
triangle right over here. If you're actually going
to take the IIT-JEE exam, I recommend having
most of the trig identities at your fingertips. But let's say that's
a right triangle. And let's call that theta. And then, this angle
right over, here is going to be pi over 2 minus
theta, if we're in radians. 90 degrees right
here is pi over 2. The whole triangle
right over here is 180 degrees, which is pi. So this is pi over 2. And so these two guys
are going to have to add up to the
other pi over 2. So this is pi over
2 minus theta. So let's think about
what cosine of theta is. Cosine of theta, "sou-cah-toa,"
cosine of theta is this side. Let's call this a, b, and c. It is going to be
the adjacent side. So cosine is adjacent. Let me write this
down, "sou-cah-toa." Cosine is adjacent
over hypotenuse. So cosine of theta is b over
c, adjacent over hypotenuse, is equal to b over c. But what's also
equal to b over c? If we look at this
angle over here, from this angle's point of
view, b is the opposite side. So it's opposite over hypotenuse
for this angle's point of view. So it's also equal to the
sine of pi over 2 minus theta. So we got our first
main identity here, that the cosine of theta
is equal to the sine of pi over 2 minus theta. And we could go the
other way around, use the exact same logic to
get that the sine of theta is equal to the cosine
of pi over 2 minus theta. So that'll be pretty
useful when we try to solve this
equation right over here. And we can even use it
when we solve this one. Because if I were to write
down the cotangent of 5 theta-- let me write this down. So the cotangent of-- not
cosine-- the cotangent of 5 theta, this is
the exact same thing. The cotangent of five theta
is the exact same thing as the cosine of 5 theta
over the sine of 5 theta. It's just 1 over the tangent. And this is the same thing. I can convert the cosine into
sine using this identity up here. This is the sine of pi
over 2 minus 5 theta over the cosine of pi
over 2 minus 5 theta. And for here, I'm just using
this identity over here. And this is the exact
same thing as the tangent of pi over 2 minus 5 theta. And since we're going
to actually have to solve this
equation eventually, and we want to make sure
we get all of the solutions to this equation. One thing that you
might want to remember is, when you take a
tangent of an angle-- and I'll just draw
the unit circle here-- so if I have some angle--
let me draw my axes. And let's say that this
right here is theta. You probably remember that
this theta is essentially the slope of the line formed. It's opposite over adjacent. It's the slope of
the line formed, or by the radius
in the unit circle. So this theta is going to
have the exact same tangent as if we add 180 degrees
or if we add pi to this. So if you add pi, you
get theta plus pi. So this whole angle right
over here is theta plus pi. And its tangent is going
to be the same thing. It's the slope of the radius
is going to be the same. So you can add multiples
of pi to a tangent value, and you're going to get
the exact same value. So let me put some
multiples of pi over here, so that
we can make sure that we get all
of the solutions. So with that said,
with this written, we can now solve this
first equation over here. And then, we can worry about
some of these other constraints later. So tangent of theta is equal
to cotangent of 5 theta. Well, so we can write tangent
of theta is equal to-- and instead of writing cotangent
of 5 theta, we can write is equal to the tangent
of pi over 2 minus 5 theta plus integer
multiples of pi. And so the tangent of this
is equal to tangent of that. These things could be
equal to each other. So we get theta is equal to pi
over 2 minus 5 theta plus n pi. Or at minimum, this
could be equal, when you take the
tangent of this, it's equal to the tangent
of either the same thing, or we could add
multiples of pi to it. You could view it as either way. But now, we just
have to solve this. We can add 5 pi to both
sides of this equation. We get, sorry, we
could add 5 theta to both sides of this equation. We're adding this to both sides. So you get 6 theta is equal
to pi over 2 minus n pi, minus multiples of pi, or
sorry, plus multiples of pi. There's a plus right over there. Divide both sides by 6. I get theta is equal to pi
over 12 plus n times pi over 6. Or just to make the
fraction adding easier, this is the same thing
as pi over 12 plus 2. Let me write it this way. Plus 2 n pi over 12. This is the same
thing as n pi over 6. I just multiplied the
numerator and denominator by 2. And this is the
exact same thing. You could imagine a 1
coefficient out here. This is the same thing as
2n plus 1 times pi over 12. So that's all of the solutions
for this first equation right over there. Now, let's do the same thing
for this second equation. And then, we'll see where
they overlap in this range. And we'll take out anything that
satisfies this criteria right over here. So the second equation. Let me write it over here. We have sine of 2 theta. Let me write it over here first. Cosine of 4 theta. What does cosine of
4 theta is equal to? Cosine of 4 theta, using
the same exact logic, is equal to sine of
pi over 2 minus theta. And of course, when we take
a sine or cosine or we take really any trig identity or
any trig function of any angle, you're going to
get the same value if you add multiples of 2 pi. So let me just add
multiples of 2 pi here, just so we can
make sure that we get all the possible solutions. So this is what cosine
of 4 theta is equal to. So let us-- oh,
let me be careful. Cosine of 4 theta is not equal
to pi over 2 minus theta. That would be cosine of theta. If it's cosine of 4 theta,
it's pi over 2 minus 4 theta plus multiples of 2 pi. So plus 2 pi n. Because obviously you
add multiples of 2 pi, you're going to go back
to the same angles. So you go back up to
this equation up here. We get sine of 2 theta is
equal to cosine of 4 theta. Which is the same thing
as this over here, which is equal to sine of
pi over 2 minus 4 theta plus 2 pi n. And so these evaluate
to the same thing. Let's set them
equal to each other. 2 theta is equal to pi over
2 minus 4 pi plus 2 pi n. Let's add 4 pi to both
sides of this equation. We get 6 theta. 6 theta is equal to pi over 2. So this is canceled out. Plus 2 pi n. Let's divide both sides by 6. We get theta is equal to
pi over 12 plus pi over 3n. 2 divided by 6 is pi n over 3. Now, we can put it over
a common denominator just to simplify things. So we have over 12. We have this pi. pi n over 3 is the same
thing as plus 4n pi over 12. Or we could write
this as being, this is equal to 4n plus
1 times pi over 12. So now, we just
need to see where there is overlap between this
solution and that solution. Remember, we just have to
count the number of solutions. We actually don't have
to find the solutions. And actually if you pretty savvy
and do things in your head, just at this point, you could
think about how many of these you're going to have
between negative pi over 2 and pi over 2. If you remember, that was the
range that we're working with. Between negative pi over 2
and pi over 2, not including those two. And then you can figure out,
and actually every one of these is also going to
be one of these. Because no matter what
you set n equal to, if you set n equal
to twice that, you're going to have
the equivalent thing. So any of these are
going to be any of these. Anything that
satisfies this equation will also satisfy this
equation right here. So really, we can just count
how many are in this one. But just to make
it clear, I'm going to find all of them that
satisfy this equation. And then we're going to
see how many of those satisfy this equation as well. But the fast way
to do it, just find the ones that satisfy this one. Or just count the ones
that satisfy this one. And you've done the problem. So let's just count. So let's just start
at n equals 0. If n is equal to 0-- and
I'm using this n up here. I won't even write that. If n is equal to 0, we
just have pi over 12. If n is 1, we get 3 pi over 12. If n is 2, we get 5 pi over 12. And we can't make n equal 3. Because if n was
equal to 3, this would be 7 pi over 12 which
is greater than pi over 2. That's greater
than 6 pi over 12. So we can't have that. We could also go to negative n. We could also have negative n. If n is negative 1, then this
becomes negative pi over 12. If n is negative 2, this
becomes negative 3 pi over 12. And if n is negative
3, then this becomes a negative 5 pi over 12. And once again, we can't go
to n is equal to negative 4, because then we
get negative 7 pi over 12 which is
out of our range. So this satisfies
this equation up here. Now, which of these are
overlap with this over here? So let's set n equal to 0. We get pi over 12. If n is equal to 1,
we get 5 pi over 12. We can't set n equal to
2, because then we'll get out of our range,
we'll go above pi over 2. If n is equal to negative 1,
we get negative 3 pi over 12. If n is equal to negative
2, this is negative 8. Then, we get to
negative 7 pi over 12, which is smaller than
negative pi over 2. So that doesn't count. So there are three solutions. Or if we look over here,
we have three solutions. We have one, two,
three solutions. And none of them satisfy this. None of them are
multiples of n pi over 5. So the answer to
our problem is 3. And you really could have just
done the second equation right over here. And realize that anything that's
a solution in this top equation would be a solution to
this bottom equation. And you would have just
counted these solutions. And you would have been done.