Question

Find the value of sin2 5o+sin210o+... Sin285o+sin290o

Answer 1

Step-by-step explanation:

Explanation:

θ can be in the first quadrant 0≤θ≤90 or the fourth quadrant 270≤θ≤360

If θ is in the first quadrant, 

then 

sinθ=513

cosθ=1213

tanθ=512

Therefore, 

sin2θ=2sinθcosθ=2×513×1213=120169

cos2θ=cos2θ−sin2θ=(1213)2−(513)2=144169−25169=119169

If θ is in the fourth quadrant, 

then

sinθ=−513

cosθ=1213

tanθ=−512

Therefore, 

sin2θ=2sinθcosθ=2×−513×1213=