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Answer 1

Answer:

1

Show that cos^2 (45°)-sin^2 (15°) = √3/4?

Anil Anand

Answered July 5, 2017

[math]\cos^2 45^\circ - \sin^2 15^\circ[/math]

[math] = \cos^2 45^\circ - \sin^2 (45^\circ - 30^\circ) \tag {LHS}[/math]

We know that [math]\sin (A - B) = sinA \cdot cosB - cosA \cdot sinB[/math]

Therefore, LHS [math]= \cos^2 45^\circ - (\sin45^\circ \cdot \cos30^\circ - \cos45^\circ \cdot \sin30^\circ)^2[/math]

[math]= \dfrac{1}{2} - \left( \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} \right)^2[/math]

[math]= \dfrac{1}{2} - \left( \dfrac{\sqrt3}{2 \sqrt2} - \dfrac{1}{2 \sqrt2} \right)^2[/math]

[math]= \dfrac{1}{2} - \left( \dfrac{\sqrt3 - 1}{2 \sqrt2} \right)[/math]

[math]=[/math] [math]\dfrac{1}{2} - \left( \dfrac{(\sqrt3 - 1)^2}{8} \right)[/math]

[math]= \dfrac{1}{2} - \dfrac{(\sqrt3 - 1)^2}{8}[/math]

[math]= \dfrac{1}{2} - \dfrac{4 - 2\sqrt3}{8}[/math]

[math]= \dfrac{1}{2} - \dfrac{2 - \sqrt3}{4}[/math]

[math]= \dfrac{2 - (2 + \sqrt3)}{4}[/math]

[math]= \dfrac{\sqrt3}{4} = [/math]RHS.

Hence, the problem is solved

Explanation:

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