If sin A -4-5 and cos B -5-13- where 0- A - B -2- find the values of the following-i sin A - B ii cos A - B iii sin A - B iv cos A - B

We have, sinA= 4/5 and B= 5/13 ∴ cosA= √(1 sin^2A) and sinB= √(1 cos^2B) ⇒ cosA= √(1 ( 4/5 )^2) and sinB= √(1 ( 5/13 )^2)sinB= √(1 ( 5/13 )^2) ⇒ cosA= √(1 16 ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

If sin A =4/5...

Question

If $s i n A = \frac{4}{5}$ and $c o s B = \frac{5}{13}$ , where $0 < A$ , $B < \frac{π}{2}$ , find the values of the following:
(i) sin(A+B)
(ii) cos(A+B)
(iii) sin(A-B)
(iv) cos(A-B)

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Solution

We have,
$s i n A = \frac{4}{5}$ and $B = \frac{5}{13}$
$∴ c o s A = \sqrt{1 - s i n^{2} A}$ and $s i n B = \sqrt{1 - c o s^{2} B}$
$\Rightarrow c o s A = \sqrt{1 - {(\frac{4}{5})}^{2}}$ and $s i n B = \sqrt{1 - {(\frac{5}{13})}^{2}}$ $s i n B = \sqrt{1 - {(\frac{5}{13})}^{2}}$
$\Rightarrow c o s A = \sqrt{1 - \frac{16}{25}}$ and $s i n B = \sqrt{1 - \frac{25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{25 - 16}{25}}$ and $s i n B = \sqrt{\frac{169 - 25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{9}{25}}$ and $s i n B = \sqrt{\frac{144}{169}} \Rightarrow c o s A = \frac{3}{5}$ and $s i n B = \frac{12}{13}$
Now,
$s i n (A + B) = s i n A c o s B + c o s A s i n B$
$= \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13} = \frac{20}{65} + \frac{36}{65} = \frac{20 + 36}{65} = \frac{56}{65}$
(ii) We have,
$s i n A = \frac{4}{5}$ and $B = \frac{5}{13}$
$∴ c o s A = \sqrt{1 - s i n^{2} A}$ and $s i n B = \sqrt{1 - c o s^{2} B}$
$\Rightarrow c o s A = \sqrt{1 {(\frac{4}{5})}^{2}}$ and
$s i n B = \sqrt{1 - {(\frac{5}{13})}^{2}}$
$\Rightarrow c o s A = \sqrt{1 - \frac{16}{15}}$
$s i n B = \sqrt{1 - \frac{25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{25 - 16}{25}}$ and
$s i n B = \sqrt{\frac{169 - 25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{9}{25}}$ and
$s i n B = \sqrt{\frac{144}{169}}$
$c o s A = \frac{3}{5}$ and
$s i n B = \frac{12}{13}$
Now,
$c o s (A + B) = c o s A c o s B - s i n A s i n B$
$= \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13}$
$= \frac{15}{65} - \frac{48}{65}$
$= \frac{15 - 48}{65} = \frac{- 33}{65}$
(iii) We have,
$s i n A = \frac{4}{5}$ and $c o s B = \frac{5}{13}$
$∴ c o s A = \sqrt{1 - s i n^{2} A}$ and $s i n B = \sqrt{1 - c o s^{2} B}$
$\Rightarrow c o s A = \sqrt{1 - {(\frac{4}{5})}^{2}}$ and
$s i n B = \sqrt{1 - {(\frac{5}{13})}^{2}}$
$\Rightarrow c o s A = \sqrt{1 - \frac{16}{25}}$ and $s i n B = \sqrt{1 - \frac{25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{25 - 16}{25}}$ and
$s i n B = \sqrt{\frac{169 - 25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{9}{25}}$ and $s i n B = \sqrt{\frac{144}{169}}$
$\Rightarrow c o s A = \frac{3}{5}$ and $s i n B = \frac{12}{13}$
Now,
$s i n (A - B) = s i n A c o s B - c o s A s i n B$
$= \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13}$
$= \frac{20}{65} + \frac{36}{65} = \frac{20 - 36}{65} = \frac{- 16}{65}$
(iv) We have,
$s i n A = \frac{4}{5}$ and $c o s B = \frac{5}{13}$
$∴ c o s A = \sqrt{1 - s i n^{2} A}$ and
$s i n B = \sqrt{1 - c o s^{2} B}$
$\Rightarrow c o s A = \sqrt{1 - {(\frac{4}{5})}^{2}}$
$s i n B = \sqrt{1 - {(\frac{5}{13})}^{2}}$
$\Rightarrow c o s A = \sqrt{1 - \frac{16}{25}}$ and
$s i n B = \sqrt{1 - \frac{25}{169}}$
$\Rightarrow c o s A = \sqrt{\frac{25 - 16}{25}}$ and
$s i n B = \frac{169 - 25}{169}$
$\Rightarrow c o s A = \sqrt{\frac{9}{25}}$ and $s i n B = \frac{12}{13}$
Now,
$c o s (A - B) = c o s A c o s B + s i n A s i n B$
$= \frac{3}{5} \times \frac{5}{13} + \frac{4}{5} \times \frac{12}{13}$
$= \frac{15}{65} + \frac{48}{65}$
$= \frac{15 + 48}{65}$
$= \frac{63}{65}$

Suggest Corrections

If sinA=45 and cosB=513, where 0<A, B<π2, find the values of the following: (i) sin(A+B) (ii) cos(A+B) (iii) sin(A-B) (iv) cos(A-B)

If $s i n A = \frac{4}{5}$ and $c o s B = \frac{5}{13}$ , where $0 < A$ , $B < \frac{π}{2}$ , find the values of the following:
(i) sin(A+B)
(ii) cos(A+B)
(iii) sin(A-B)
(iv) cos(A-B)