Question

can anyone answer this one(8)

Answer 1

Answer:

sin A = cos(90-A)

Step-by-step explanation:

this is because

sin and cos both are complementary angles

We want to prove that the sine of an angle equals the cosine of its complement.

\sin(\theta) = \cos(90^\circ-\theta)sin(θ)=cos(90  

∘

−θ)

[I'm skeptical. Please show me an example.]

\sin(10^\circ)  

\cos(80^\circ)  

 

Let's start with a right triangle. Notice how the acute angles are complementary, sum to 90

sin(\theta)sin(θ) and \cos(90-theta)cos(90  

∘

−θ), give the exact same side ratio in a right triangle.

And we're done! We've shown that \sin(\theta) = \cos(90^\circ-\theta)sin(θ)=cos(90  

∘

−θ).

In other words, the sine of an angle equals the cosine of its complement.

Well, technically we've only shown this for angles between 0^\circ  

 and 90^

. To make our proof work for all angles, we'd need to move beyond right triangle trigonometry into the world of unit circle trigonometry, but that's a task for another time