215k views
1 vote
Sin 70° + sin50°= 3 cos 10°

Options:

(A) 100​
(B) 200​
(C) 300​
(D) 400​

1 Answer

0 votes

Final answer:

Using trigonometric identities, sin 70° + sin 50° simplifies to √3 cos(10°), which is not equal to 3 cos 10°. Therefore, the original equation is not true, and none of the provided options are correct.

Step-by-step explanation:

To solve the given problem, we will use the sum-to-product trigonometric identities. Specifically, the identity for the sum of two sine functions:

sin A + sin B = 2 sin((A + B)/2) cos((A - B)/2).

Applying this identity to sin 70° + sin 50°:

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

= 2 sin(60°) cos(10°)

Now, using the fact that sin(60°) = √3/2, the equation becomes:

= 2 (√3/2) cos(10°)

= √3 cos(10°)

The original problem was:

sin 70° + sin 50° = 3 cos 10°

If we replace the left side of the equation with what we found, we get:

√3 cos(10°) = 3 cos 10°

To solve for √3, we divide both sides by cos(10°):

√3 = 3

Since we know that √3 is not equal to 3, we can conclude that the original statement is not true with any of the provided options (A) 100, (B) 200, (C) 300, or (D) 400.

User Msamardzic
by
7.6k points