If (A − B) = A B − A B and (A − B) = + A B, find the values of sin 15° and cos 15°.

Given: (A-B)= - (1) (A-B)= + (2) To find: The values of sin 15° and cos 15° In this problem we need to find sin 15 and cos 15° Hence to get 15° angle we need to choose the value of A and B such that (A - B) = 15° So if we choose A = 45° and B = 30° Then we get, (A - B) = 15° Therefore by substituting A = 45° and B = 30° in equation (1) We get, (45^ -30^ )= 45^ 30^ - 45^ 30^ Therefore, (15^ )= 45^ 30^ - 45^ 30^ (3) Now we know that, 45^ = 45^ =1 2 , 30^ =1 2 , 30^ = 3 2 Now by substituting above values in equation (3) We get, (15^ )= (1 2 ) ( 3 2 )- (1 2 ) (1 2 ) = 3 22 -1 22 = 3-1 22 Therefore, (15^ )= 3-1 22 ....(4) Now by substituting A = 45° and B = 30° in equation (2) We get, (45^ -30^ )= 45^ 30^ + 45^ 30^ Therefore, (15^ )= 45^ 30^ + 45^ 30^ (5) Now we know that, 45^ = 45^ =1 2 , 30^ =1 2 , 30^ = 3 2 Now by substituting above values in equation (5) We get, (15^ )= (1 2 ) ( 3 2 )+ (1 2 ) (1 2 ) = 3 22 +1 22 = 3+1 22 Therefore, (15^ )= 3+1 22 (6) Therefore from equation (4) and (6) (15^ )= 3-1 22 (15^ )= 3+1 22

If (A − B) = A B − A B and (A − B) = + A B, find the values of si…

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Question

If $\sin (\text{A} − \text{B}) = \sin \text{A} \cos \text{B} − \cos \text{A} \sin \text{B}$ and $\cos (\text{A − B}) = \cos\text{A} \cos\text{B} + \sin \text{A} \sin \text{B},$ find the values of sin 15° and cos 15°.

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Answer

Given:
$\sin(\text {A}-\text{B})=\sin\text {A}\cos\text {B}-\cos\text{A}\sin\text {B}\ \dots(1)$
$\cos(\text {A}-\text{B})=\cos\text {A}\cos\text {B}+\sin\text{A}\sin\text {B}\ \dots(2)$
To find: The values of sin 15° and cos 15°
In this problem we need to find sin 15 and cos 15°
Hence to get 15° angle we need to choose the value of A and B such that (A - B) = 15°
So if we choose A = 45° and B = 30°
Then we get, (A - B) = 15°
Therefore by substituting A = 45° and B = 30° in equation (1)
We get,
$\sin(45^\circ-30^\circ)=\sin45^\circ\cos30^\circ-\cos45^ \circ\sin30^\circ$
Therefore,
$\sin(15^\circ)= \sin45^\circ\cos30^\circ-\cos45^\circ \sin30^\circ\ \dots(3)$
Now we know that,
$\sin45^\circ=\cos45^ \circ=\frac{1}{\sqrt{2}},\ \sin30^ \circ=\frac{1}{2},\ \cos30^\circ= \frac{\sqrt{3}}{2}$
Now by substituting above values in equation (3)
We get,
$\sin(15^\circ)=\bigg (\frac{1}{\sqrt{2}}\bigg)\times\bigg (\frac{\sqrt{3}}{2}\bigg)-\bigg(\frac {1}{\sqrt{2}}\bigg)\times\bigg(\frac{1} {2}\bigg)$
$=\frac{\sqrt{3}} {2\sqrt{2}}-\frac{1}{2\sqrt{2}}$
$=\frac{\sqrt{3}-1} {2\sqrt{2}}$
Therefore,
$\sin(15^\circ)=\frac {\sqrt{3}-1}{2\sqrt{2}}\ ....(4)$
Now by substituting A = 45° and B = 30° in equation (2)
We get,
$\cos(45^\circ-30^ \circ)=\cos45^\circ\cos30^\circ+\sin45^ \circ\sin30^\circ$
Therefore,
$\cos(15^\circ)= \cos45^\circ\cos30^\circ+\sin45^\circ \sin30^\circ\ \dots(5)$
Now we know that,
$\sin45^\circ=\cos45^ \circ=\frac{1}{\sqrt{2}},\ \sin30^ \circ=\frac{1}{2},\ \cos30^\circ=\frac {\sqrt{3}}{2}$
Now by substituting above values in equation (5)
We get,
$\cos(15^\circ)=\bigg (\frac{1}{\sqrt{2}}\bigg)\times\bigg (\frac{\sqrt{3}}{2}\bigg)+\bigg(\frac {1}{\sqrt{2}}\bigg)\times\bigg(\frac{1} {2}\bigg)$
$=\frac{\sqrt{3}} {2\sqrt{2}}+\frac{1}{2\sqrt{2}}$
$=\frac{\sqrt{3}+1} {2\sqrt{2}}$
Therefore,
$\cos(15^\circ)=\frac {\sqrt{3}+1}{2\sqrt{2}}\ \dots(6)$
Therefore from equation (4) and (6)
$\sin(15^\circ)=\frac {\sqrt{3}-1}{2\sqrt{2}}$
$\cos(15^\circ)=\frac {\sqrt{3}+1}{2\sqrt{2}}$

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