If for some θ lying between 0° and 90°, tanθ = 1, then what is the value of sin 2θ - 2sinθ cosθ ?
-1/2
Given:
tanθ = 1; 0º < θ < 90º
Formula used:
tan 45º = 1
sin 45º = 1/√2
cos 45º = 1/√2
Calculation:
We have, tanθ = 1
⇒ tanθ = tan45°
⇒ θ = 45º
Now, we need to find the value of sin2θ - 2sinθ cosθ when θ = 45º
⇒ sin245º - 2 sin45º cos45º
⇒ (1/2) - 2(1/√2)(1/√2)
∴ sin2θ - 2sinθ cosθ = - 1/2
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