Question

prove that
Cosze = cos²o-sino For OER.​

Answer 1

EXPLANATION.

Prove that.

⇒ cos2θ = cos²θ - sin²θ.

As we know that,

Formula of :

⇒ cos(A + B) = cos(A).cos(B) - sin(A).sin(B).

Using this formula in equation, we get.

⇒ cos(θ + θ) = cos(θ).cos(θ) - sin(θ).sin(θ).

⇒ cos(θ + θ) = cos²θ - sin²θ.

Hence proved.

                                                                                                                         

MORE INFORMATION.

Trigonometric ratios of multiple angles.

(1) = sin2θ =2sinθcosθ = 2tanθ/1 + tan²θ.

(2) = cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - +2sin²θ = 1 - tan²θ/1 + tan²θ.

(3) = tan2θ = 2tanθ/1 - tan²θ.

(4) = sin3θ = 3sinθ - 4sin³θ.

(5) = cos3θ = 4cos³θ - 3cosθ.

(6) = tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.

Answer 2

Step-by-step explanation:

ANSWER ✍️

Prove that.

⇒ cos2θ = cos²θ - sin²θ.

As we know that,

Formula of :

⇒ cos(A + B) = cos(A).cos(B) - sin(A).sin(B).

Using this formula in equation, we get.

⇒ cos(θ + θ) = cos(θ).cos(θ) - sin(θ).sin(θ).

⇒ cos(θ + θ) = cos²θ - sin²θ.

Hence proved.

                                                                                                                         

MORE INFORMATION.

Trigonometric ratios of multiple angles.

(1) = sin2θ =2sinθcosθ = 2tanθ/1 + tan²θ.

(2) = cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - +2sin²θ = 1 - tan²θ/1 + tan²θ.

(3) = tan2θ = 2tanθ/1 - tan²θ.

(4) = sin3θ = 3sinθ - 4sin³θ.

(5) = cos3θ = 4cos³θ - 3cosθ.

(6) = tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.

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