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### Course: Class 10 (Old) > Unit 8

Lesson 4: Trigonometric ratios of complementary angles# Sine & cosine of complementary angles

Sal shows that the sine of any angle is equal to the cosine of its complementary angle. Created by Sal Khan.

## Want to join the conversation?

- I tried the operation on a calculator and found that:

sin(32) is about 0.55

cos(58) is about 0.12

What am i missing?(14 votes)- Your calculator needs to be set in degree mode. Then they will be equivalent measurements.(48 votes)

- Just to clarify an SAT question, is sin(xº) always equal to cos(90º-xº)?

*****SPOILER: below is one of the Khan academy practice test questions*****

In a right triangle, one angle measures xº where sin(xº) = 4/5. What is cos(90º - xº)?(14 votes)- Well I'm answering it 4 years after it has been asked but anyways,...

we know that, sin(x)=cos(90-x)

and we have sin(x)=4/5

so we can deduce from above equations that

cos(90-x)=4/5.. hope it helps..(22 votes)

- What is an arbitrary angle?(10 votes)
- That means what he's saying applies to any angle regardless of measure.(24 votes)

- can someone please help me i feel dumb :((12 votes)
- Well, we know that the sine of an angle is the ratio of the opposite to hypotenuse. Similarly, the cosine of an angle is the ratio of the adjacent side to the hypotenuse. If we pause and imagine a right triangle, the sine of one angle would be the cosine of the angle across from it, since the hypotenuse is constant, but the opposite side of one angle and the adjacent side of the other angle refer to the same side. Since we are talking about a right triangle, the angles are complementary. And this fact gives us enough information to conclude the following equation:

sin(x degrees) = cos(90 - x degrees), and vice versa.

If you have any further questions, please leave them in a comment, and I'll get right to them!(18 votes)

- What is near that 90° - ø thing?

It's the (ø) that triggers me.(5 votes)- It's actually θ, the Greek letter theta. Lowercase greek letters are commonly used to represent angle measures. You might also see alpha, which looks like an a, as well as many others. So that you don't get lost, here is a copy of the Greek alphabet: ςερτυθιοπασδφγηξκλζχψωβνμ.

If you look closely, you'll see that π is in there, since it is also a Greek letter.

Don't get too confused, these work the same way as any other variable, like a, b, x, and y.

Hope that helps!(26 votes)

- What is the difference between radians and degrees?(7 votes)
- They're just different units to measure the same quantity (angle measure), like how pounds and kilograms are different units for the same quantiy (mass). A degree is defined so that there are 360 degrees in a full circle, and a radian is defined so that there are 2π radians in a full circle.(21 votes)

- I have a question on this concept:

If sin(x) = cos(90 - x)

then how do I find the x in

sin(50 - x) = cos(3x + 10)(6 votes)- You could rearrange the concept a bit to get that the sum of the arguments must be 90 degrees for the sides to be equal, since the sine is the same as the cosine of the complementary angle. We can then set up an equation with just the arguments:

50 - x + 3x + 10 = 90

2x + 60 = 90

2x = 30

x = 15(14 votes)

- im gonna be honest i dont understand a single thing i don't even understand how he did it(5 votes)
- If you think about it in terms of a right triangle, you can have angles and opposite sides, let C be the right angle and c be the hypotenuse. Then you have angle A and side opposite a and angle B and side opposite b. The sin(A)=opp/hyp ]=a/c and the cos(A)=adj/hyp=b/c. Also, the sin(B)=b/c and cos(B)=a/c. We also know that with a right triangle, the two acute angles have to add up to 90 degrees (or complementary) M<A+M<B=90 (thus m<A=90-m<B and m<B-=90-m<A).. Looking at this, sin(A)=cos(B) and cos(A)=sin(B), but with substitution, you could also say sin(A)=cos(90-A) and cos(A)=sin(90-A).(5 votes)

- This is a really cool concept! But how does this help you solve a triangle? I mean what sort of question would this work on?(4 votes)
- Hello, 8 years late reply here. You probably already have the answer, but it can help in finding triangle sides or angles, I suppose...(2 votes)

- I am completely new to this topic. Can someone tell me where to start? I currently am in Algebra I.(2 votes)
- Trigonometry is generally learnt after finishing Algebra. Khan Academy's Algebra 1 courses are at https://khanacademy.org/math/algebra

After you finish Algebra 1 and Algebra 2 you'll move on to Trigonometry, which is what this video is covering :)(3 votes)

## Video transcript

We see in a triangle, or I
guess we know in a triangle, there's three angles. And if we're talking
about a right triangle, like the one that I've
drawn here, one of them is going to be a right angle. And so we have two other
angles to deal with. And what I want to
explore in this video is the relationship between
the sine of one of these angles and the cosine of the other, the
cosine of one of these angles and the sine of the other. So to do that, let's just
say that this angle-- I guess we could call
it angle A-- let's say it's equal to theta. If this is equal to
theta, if it's measure is equal to theta degrees, say,
what is the measure of angle B going to be? Well, the thing that
will jump out at you-- and we've looked at
this in other problems-- is the sum of the
angles of a triangle are going to be 180 degrees. And this one right over
here, it's a right triangle. So this right angle takes
up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to
have to add up to 90 degrees. This one and this one,
angle A and angle B, are going to be
complements of each other. They're going to
be complementary. Or another way of
thinking about it is B could be written
as 90 minus theta. If you add theta to 90
degrees minus theta, you're going to get 90 degrees. Now, why is this interesting? Well, let's think about what
the sine of theta is equal to. Sine is opposite
over hypotenuse. The opposite side is BC. So this is going
to be the length of BC over the hypotenuse. The hypotenuse is side AB. So the length of BC
over the length of AB. Now, what is that
ratio if we were to look at this angle
right over here? Well, for angle B, BC
is the adjacent side, and AB is the hypotenuse. From angle B's
perspective, this is the adjacent over
the hypotenuse. Now, what trig ratio is
adjacent over hypotenuse? Well, that's cosine. Sohcahtoa, let me
write that down. Doesn't hurt. Sine is opposite
over hypotenuse. We see that right over there. Cosine is adjacent
over hypotenuse, cah. And toa, tangent is
opposite over adjacent. So from this
angle's perspective, taking the length of BC,
BC is its adjacent side, and the hypotenuse is still AB. So from this
angle's perspective, this is adjacent
over hypotenuse. Or another way of
thinking about it, it's the cosine of this angle. So that's going to be
equal to the cosine of 90 degrees minus theta. That's a pretty
neat relationship. The sine of an angle is equal
to the cosine of its complement. So one way to think about
it, the sine of-- we could just pick any
arbitrary angle-- let's say, the sine of 60
degrees is going to be equal to the cosine of what? And I encourage you to pause
the video and think about it. Well, it's going to be
the cosine of 90 minus 60. It's going to be the
cosine of 30 degrees. 30 plus 60 is 90. And of course, you could
go the other way around. We could think about
the cosine of theta. The cosine of theta is going to
be equal to the adjacent side to theta, to angle
A, I should say. And so the adjacent
side is right over here. That's AC. So it's going to be AC
over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from
angle B's point of view? Well, the sine of
angle B is going to be its opposite side,
AC, over the hypotenuse, AB. So this right over here,
from angle B's perspective, this is angle B's sine. So this is equal to the sine
of 90 degrees minus theta. So the cosine of an angle
is equal to the sine of its complement. The sine of an angle is equal
to the cosine of its complement.