Prove the following trig identity: 1 _ sin? x _ 2cosx cos- x cosx 1 +secx [4
The Correct Answer and Explanation is:
To prove a trigonometric identity, we aim to simplify one side of the equation and show that it equals the other. However, your expression appears to be incomplete or improperly formatted. From your message:
1 _ sin? x _ 2cosx cos- x cosx 1 +secx [4
This seems to be a corrupted or unclear identity. Based on common trigonometric identities, I will try to interpret it as:
(1 − sin²x) − 2cosx * cos(−x) * cosx = 1 + secx
Let’s clarify the terms and rewrite the left-hand side (LHS) using known identities.
Step 1: Use Pythagorean Identity
Recall that:
- 1 − sin²x = cos²x
- cos(−x) = cosx (cosine is even)
So we simplify the LHS:
LHS = (1 − sin²x) − 2cosx * cos(−x) * cosx
= cos²x − 2cosx * cosx * cosx
= cos²x − 2cos³x
Step 2: Factor the expression
Now factor the expression:
LHS = cos²x − 2cos³x = cos²x(1 − 2cosx)
Step 3: Try to Simplify or Compare with RHS
The RHS of your expression is 1 + secx, where:
secx = 1/cosx
So,
RHS = 1 + 1/cosx
This is not obviously equal to cos²x(1 − 2cosx). That suggests either:
- There was a mistake in transcribing the original identity, or
- The two sides of the equation are not equal, hence not an identity.
Final Notes
Based on what you provided, the left-hand side simplifies to cos²x(1 − 2cosx), while the right-hand side simplifies to 1 + 1/cosx. These expressions are not algebraically equivalent, so as written, the identity cannot be proven because the two sides are not equal.
