Question

If sin(A+B) = sinA cosB + cosA sinB, find the value of sin75 and sin105.

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Answer 1

Answer:

just substitute 45+30 in place of A+B and then find the value using above identity

similary for 105 you can write it as 60+ 45 in place of A+B

Answer 2

Given:,

• sin(A + B) = sin A cos B + cos A sin B

Find:

• sin 75°
• sin 105°

Solution:

we, can write

$\sf \dashrightarrow \sin {75}^{ \circ} = \sin{(45 + 30)}^{ \circ}$

Then,

$\sf \dashrightarrow \sin{(45 + 30)}^{ \circ} = \sin {45}^{ \circ} \cos {30}^{ \circ} + \cos {45}^{ \circ} \sin {30}^{ \circ} \\ \\ \\ \sf where, \sin {45}^{ \circ} = \dfrac{1}{ \sqrt{2} } \\ \cos {30}^{ \circ} = \dfrac{ \sqrt{3} }{2} \\ \cos {45}^{ \circ} = \dfrac{1}{ \sqrt{2} } \\ \sin {30}^{ \circ} = \dfrac{1}{2} \\ \qquad \qquad \sf \dag using \: these \: values \dag \\ \\ \sf \dashrightarrow \sin{(45 + 30)}^{ \circ} = \bigg( \dfrac{1}{ \sqrt{2} } \bigg) \bigg( \dfrac{ \sqrt{3} }{ 2 } \bigg) + \bigg( \dfrac{1}{ \sqrt{2} } \bigg) \bigg( \dfrac{1}{2} \bigg) \\ \\ \\ \sf \dashrightarrow \sin{(45 + 30)}^{ \circ} = \bigg( \dfrac{ \sqrt{3} }{ 2\sqrt{2} } \bigg)+ \bigg( \dfrac{1}{ 2\sqrt{2} } \bigg) \\ \\ \\ \sf \dashrightarrow \sin{(75)}^{ \circ} = \dfrac{ \sqrt{3} }{ 2\sqrt{2} } + \dfrac{1}{ 2\sqrt{2} } \\ \\ \\ \sf \dashrightarrow \sin{(75)}^{ \circ} = \dfrac{ \sqrt{3} + 1}{ 2\sqrt{2} }\\ \\ \\ \sf \therefore \sin{75}^{ \circ} = \dfrac{ \sqrt{3} + 1}{ 2\sqrt{2} }$

we, know that

sin 105° and sin 75° are supplementary angles. Hence, Both have same values.

$\sf \implies \sin {105}^{\circ} = \sin {75}^{\circ}$

where,

sin 75° = $\dfrac{ \sqrt{3} + 1}{ 2\sqrt{2} }$

So,

sin 105° = $\dfrac{ \sqrt{3} + 1}{ 2\sqrt{2} }$

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Hence,

• sin 75° = $\dfrac{ \sqrt{3} + 1}{ 2\sqrt{2} }$

• sin 105° = $\dfrac{ \sqrt{3} + 1}{ 2\sqrt{2} }$

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