Question

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find the value of
sin15 degree +cos15 degree

Answer 1

To Find :

The value of :

sin15° + cos15°

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We know that :

• sin15° = sin(45° -30°)
• cos15° = cos(45°-30°)

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Formula To be used :

The formula of compound trigonometric angles :

• sin(A-B) = sinA.cosB - cosA.sinB
• Cos(A-B) = cosA.cosB + sinA.sinB

Here, A = 45°

B = 30°

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Solution :

we know that,

Sin15° = sin(45°-30°)

$\tt{ \implies sin15 \degree = sin45 \degree \times cos 30 \degree + cos45 \degree \times sin30 \degree}$

$\tt{ \implies \: sin15 \degree = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } \times \frac{1}{2} }$

$\tt{ \implies \: sin15 \degree = \frac{ \sqrt{3} }{ 2\sqrt{2} } - \frac{1}{2 \sqrt{2} } }$

$\tt{ \implies \: sin15 \degree = \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } }$

Cos 15° = cos(45°-30°)

$\tt{ \implies \: cos15 \degree = cos45 \degree \times cos 30 \degree + sin45 \degree \times sin30 \degree}$

$\tt{ \implies \: cos15 \degree = \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } \times \frac{1}{2} }$

$\tt{ \implies \: cos15 \degree = \frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} } }$

$\implies \tt{ \: cos15 \degree = \frac{ \sqrt{3} + 1}{2 \sqrt{2} } }$

so, we have got the value of sin15° and cos15° .

so, sin15°+cos15° :

$\tt{ \implies \: sin15 \degree + cos15 \degree = \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } + \frac{ \sqrt{3} + 1}{2 \sqrt{2} } }$

$\tt{\implies \: sin 15 \degree + cos15 \degree = \frac{( \sqrt{3} - 1) + ( \sqrt{3} + 1)}{2 \sqrt{2} } }$

$\tt{ \implies \: sin15 \degree + cos 15 \degree = \frac{2 \sqrt{3} }{2 \sqrt{2} } }$

$\tt{ \implies \: sin15 \degree + cos15 \degree = \frac{ \sqrt{3} }{ \sqrt{2} } }$

$\red{ \underline{ \boxed{ \tt{ \implies \: sin15 \degree + cos15 \degree = \sqrt{ \frac{3}{2} } }}}}$

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Additional Information :

1] Trigonometric Ratio of compound Angles :

$\tt{ \bull \: sin (a + b) = sin \: a \times cos \: b + cos \: a \times sin \: b}$

$\tt{ \bull \: sin(a - b) = sin \: a \times cos \: b - sin \: b \times \: cos \: a }$

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