Write sin 61° in terms of cosine

Answer 1

If you draw a right triangle ABC where C is the 90 degree angle.

m∠A+m∠B+m∠C=180°

m∠A+m∠B+90°=180°

m∠A+m∠B=90°

m∠B=90°-m∠A

So if

m∠A=61°

m∠B=29°

Now label the sides, the one opposite from ∠A is "a", opposite from ∠B is "b" and opposite of ∠C is "c" (the hypotenuse.

sinA=a/c

sinB=b/c

cosA=b/c

cosB=a/c

So sinA=cosB

sin61=cos29

Basically sinx=cos(90-x) and cosx=sin(90-x)