about complementary angles.. (1 Viewer)

[hitomi]

can somebody explain me this..

sin20+cos70

i dont really know how to do it..
i know its easy but im really confused,

 

[Forbidden.]

I will assume they are in degrees.

Remember that cos θ = sin (90 - θ) ?
So sin 20 + cos 70 = sin 20 + sin (90 - 70) = sin 20 + sin 20.
however sin 20 + sin 20 is not equal to sin 40.

There is a trigonometric identity generally used where:
sin u + sin v = 2 sin (u + v / 2) cos (u - v / 2)
So let u = 20 and v = 20
sin 20 + sin 20 = 2 sin (20 + 20 / 2) cos (20 - 20 / 2) = 2 sin 20

Or simply:

sin θ + sinθ = 2 sin θ

i.e.
sin 20 + sin 20 = 2 sin 20

 

[hitomi]

I will assume they are in degrees.

Remember that cos θ = sin (90 - θ) ?
So sin 20 + cos 70 = sin 20 + sin (90 - 70) = sin 20 + sin 20.
however sin 20 + sin 20 is not equal to sin 40.

There is a trigonometric identity generally used where:
sin u + sin v = 2 sin (u + v / 2) cos (u - v / 2)
So let u = 20 and v = 20
sin 20 + sin 20 = 2 sin (20 + 20 / 2) cos (20 - 20 / 2) = 2 sin 20

Or simply:

sin θ + sinθ = 2 sin θ

i.e.
sin 20 + sin 20 = 2 sin 20

thanks a lot!!
that really really help me..

 

[hitomi]

im just wondering why is it
cos θ = sin (90 - θ)?

can it be
sin θ = cos(90-θ)?

 

[cwag]

yes..both work

 

[vds700]

im just wondering why is it
cos θ = sin (90 - θ)?

can it be
sin θ = cos(90-θ)?

they're quite easy to prove. Just draw a right-angled triangle with one angle θ and the other angle will be (90 - θ) (angle sum of triangle = 180). You will see that sinθ = cos(90-θ) and cosθ = sin(90-θ).