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Math  /  Trigonometry

QuestionSin⁡2αCos⁡2β−Cos⁡2αSin⁡2β=Sin⁡2α−Sin⁡2β\operatorname{Sin}^{2} \alpha \operatorname{Cos}^{2} \beta-\operatorname{Cos}^{2} \alpha \operatorname{Sin}^{2} \beta=\operatorname{Sin}^{2} \alpha-\operatorname{Sin}^{2} \beta

Studdy Solution

STEP 1

1. The equation involves trigonometric functions: sin⁡\sin and cos⁡\cos.
2. The equation can be simplified using trigonometric identities.
3. The goal is to verify the given equation's correctness by manipulation and simplification.

STEP 2

1. Expand and simplify the left-hand side (LHS) of the equation.
2. Compare the simplified form of the LHS with the right-hand side (RHS) of the equation.

STEP 3

Start with the left-hand side (LHS) of the equation:
sin⁡2αcos⁡2β−cos⁡2αsin⁡2β\sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta

STEP 4

Use the trigonometric identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x to rewrite sin⁡2α\sin^2 \alpha and sin⁡2β\sin^2 \beta in terms of cos⁡2α\cos^2 \alpha and cos⁡2β\cos^2 \beta:
sin⁡2α=1−cos⁡2αandsin⁡2β=1−cos⁡2β\sin^2 \alpha = 1 - \cos^2 \alpha \quad \text{and} \quad \sin^2 \beta = 1 - \cos^2 \beta

STEP 5

Substitute sin⁡2α=1−cos⁡2α\sin^2 \alpha = 1 - \cos^2 \alpha and sin⁡2β=1−cos⁡2β\sin^2 \beta = 1 - \cos^2 \beta into the LHS:
(1−cos⁡2α)cos⁡2β−cos⁡2α(1−cos⁡2β)(1 - \cos^2 \alpha) \cos^2 \beta - \cos^2 \alpha (1 - \cos^2 \beta)

STEP 6

Distribute the terms inside the parentheses:
cos⁡2β−cos⁡2αcos⁡2β−cos⁡2α+cos⁡2αcos⁡2β\cos^2 \beta - \cos^2 \alpha \cos^2 \beta - \cos^2 \alpha + \cos^2 \alpha \cos^2 \beta

STEP 7

Combine like terms:
cos⁡2β−cos⁡2α\cos^2 \beta - \cos^2 \alpha

STEP 8

Consider the right-hand side (RHS) of the equation:
sin⁡2α−sin⁡2β\sin^2 \alpha - \sin^2 \beta

STEP 9

Substitute sin⁡2α=1−cos⁡2α\sin^2 \alpha = 1 - \cos^2 \alpha and sin⁡2β=1−cos⁡2β\sin^2 \beta = 1 - \cos^2 \beta into the RHS:
(1−cos⁡2α)−(1−cos⁡2β)(1 - \cos^2 \alpha) - (1 - \cos^2 \beta)

STEP 10

Simplify the expression:
1−cos⁡2α−1+cos⁡2β1 - \cos^2 \alpha - 1 + \cos^2 \beta
cos⁡2β−cos⁡2α\cos^2 \beta - \cos^2 \alpha

STEP 11

Compare the simplified forms of the LHS and RHS. Both are:
cos⁡2β−cos⁡2α\cos^2 \beta - \cos^2 \alpha
Solution: Since the simplified forms of both the LHS and RHS are equal, the original equation is verified to be correct:
sin⁡2αcos⁡2β−cos⁡2αsin⁡2β=sin⁡2α−sin⁡2β\sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta = \sin^2 \alpha - \sin^2 \beta

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