Show that sin70°+sin50°= √3cos0°

Answer 1

We will use the sum-to-product identities to prove this statement.

sin(70°) + sin(50°) = 2sin((70° + 50°)/2)cos((70° - 50°)/2)
= 2sin(60°)cos(10°)

We know that sin(60°) = √3/2 and cos(10°) = (√3 + 1)/2.

Therefore,
2sin(60°)cos(10°) = 2 * √3/2 * (√3 + 1)/2
= √3 * (√3 + 1)
= (√3)^2 + √3
= 3 + √3.

So, sin(70°) + sin(50°) = 3 + √3.

Now, let's calculate the value of √3cos(0°):

√3cos(0°) = √3 * 1
= √3.

Therefore, sin(70°) + sin(50°) does not equal √3cos(0°).