What is Cos²A - Cos²B = ?

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What is Cos²A - Cos²B = ?

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Answer 1 · 2018-02-16T15:47:16

$= \frac{1}{2} (cos 2 A - cos 2 B)$ $=1/2(cos2A-cos2B)$

Explanation:

${cos}^{2} A = \frac{1}{2} (1 + cos 2 A)$ $cos^2A=1/2(1+cos2A)$
${cos}^{2} B = \frac{1}{2} (1 + cos 2 B)$ $cos^2B=1/2(1+cos2B)$

Thus,

${cos}^{2} A - {cos}^{2} B = \frac{1}{2} (1 + cos 2 A) - \frac{1}{2} (1 + cos 2 B)$ $cos^2A-cos^2B=1/2(1+cos2A)-1/2(1+cos2B)$

$= \frac{1}{2} (1 + cos 2 A - 1 - cos 2 B)$ $=1/2(1+cos2A-1-cos2B)$
$= \frac{1}{2} (cos 2 A - cos 2 B)$ $=1/2(cos2A-cos2B)$

Answer 2 · 2018-02-16T15:54:26

-sin(A+B)sin(A-B)

Explanation:

${cos}^{2} A - {cos}^{2} B = (cos A + cos B) (cos A - cos B)$ $cos^2A-cos^2B=(cosA+cosB)(cosA-cosB)$ ---------- $1$ $color(red)(1)$
Now, $cos A - cos B = 2 sin (A + B) / 2 \times sin (B - A) / 2$ $cosA-cosB=2sin(A+B)//2 xx sin(B-A)//2$
Or, $cos A - cos B = - 2 sin (A + B) / 2 \times sin (A - B) / 2$ $cosA-cosB=-2sin(A+B)//2 xx sin(A-B)//2$

Also, $cos A + cos B = 2 cos (A + B) / 2 \times cos (A - B) / 2$ $cosA+cosB=2cos(A+B)//2xxcos(A-B)//2$

Plugging in the values in $1$ $color(red)(1)$ ,and rearranging,we get,

${- 2 cos (A + B) / 2 \times sin (A + B) / 2} \times {2 sin (A - B) / 2 \times cos (A - B) / 2}$ ${-2cos(A+B)//2xxsin(A+B)//2} xx {2sin(A-B)//2 xx cos(A-B)//2}$
Thus,applying the formula $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ ,we have,
${cos}^{2} A - {cos}^{2} B = - sin (A + B) sin (A - B)$ $cos^2A-cos^2B=-sin(A+B)sin(A-B)$

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What is Cos²A - Cos²B = ?

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