how does simplifying and evaluating sin^2 20˚ + sin^2 70˚ equal to one?

Question

jhonyy9 · Answer

sin^2 20° = (0,34)^2
sin^2 70° = 1 - cos^2 70° = 1 - (0,34)^2 

in this way than we rewrite sin^2 20° + sin^2 70° = sin^2 20° +1 - cos^2 70° =
= (0,34)^2 +1 - (0,34)^2 = 1

hope helped

jhonyy9 · Answer

@ganeshie8     your opinion please 
@TheSmartOne 

ty.

pythagoras123 · Answer

This can be proven algebraically without finding the exact value of either trigonometric function. 

$$\sin^220^0 + \sin^270^0 = \sin^220^0-\cos^270^0+1$$
$$=(\sin20^0-\cos70^0)(\sin20^0 + \cos20^0)+1$$ 

Let sin20 be x. We can construct a triangle to find the relation between sin20 and cos70:

|dw:1472202216780:dw| 

Since cosx=adjacent / hypotenuse, cos70 = x / 1 = x = sin20. 

Hence our expression simplifies to 

$$0(\sin20^0+\cos20^0)+1$$

$$=1$$ (answer)

unklerhaukus · Answer

$$\sin(\theta)=\cos(90°-\theta)$$$$\sin^2(\theta)+\cos^2(\theta)=1$$

divyankasc · Answer

sin(θ)=cos(90°−θ)
sin2(θ)+cos2(θ)=1