Prove that: cos³20° +sin³10° = 3/4 (cos20°-sin10°) | Values of Trigonometric Ratios | SciPiPupil

Prove that: $cos^320° +sin^310° = \frac{3}{4}(cos20°+sin10°)$

Solution:

We know,

$cos3A = 4cos^3A-3cosA$

or, $4cos^3A = cos3A +3cosA$

or, $cos^3A = \frac{cos3A+3cosA}{4}$

Taking A = 20°

$cos^320° = \frac{cos60°+3cos20°}{4} \;--(i)$

Also,

$sin3B = 3sinB -4sin^3B$

or, $4sin^3B = 3sinB -sin3B$

or, $sin^3B = \frac{3sinB- sin3B}{4}$

Taking B = 10°

$sin^310° = \frac{3sin10°-sin30°}{4}\;--(ii)$

Adding (i) and (ii), we get;

$cos^320°+sin^310° =$

 $\dfrac{cos60°+3cos20°}{4} +\dfrac{3sin10°-sin30°}{4}$

or, 

$cos^320°+sin^310° = $

$\dfrac{cos60°+3cos20°+ 3sin10°-sin30°}{4}$

or, 

$cos^320°+sin^310° =$

 $\dfrac{sin30°+3cos20°+ 3sin10°-sin30°}{4}$

$\therefore cos^320°+sin^310° =$

$\dfrac{3}{4}(cos20°-sin10°)$

Explanation to this solution:

We know,

cos³20°+sin³10° = (cos20°+sin10°)(cos²20°+sin²10°-cos20°sin10°)

We get this using the formula, (a³+b³) = (a+b)(a²-ab+b²)

But, we can go no further in simple manner because the value of cos20 and sin10 are not known from the value table. 

However, if we use the multiple angle formula of cos3A and sin3A, we can find the solution to this answer.

Here is the Facebook link to the solution of this question in image. 

Related Notes:

Link: Introduction To Trigonometry

Link: Values of Trigonometric Ratios

Link: Compound Angles

Question: Prove that: cos³20° +sin³10° = 3/4 (cos20°-sin10°) | Values of Trigonometric Ratios | SciPiPupil

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