evaluate cos^2 (pi/4+x)-sin^2 (pi/4-x)
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=0
Step-by-step explanation:
cos²(π/4+x)-sin²(π/4-x)
=(cos(π/4+x))²-(sin(π/4-x))²
=(cos(π/4)cosx -sin(π/4)sinx)² - (sin(π/4)cosx -cos(π/4)sinx)²
=(1/√2 (cosx-sinx))² -(1/√2 (cosx-sinx))²
=0
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