Question

If $\sin A=\dfrac{4}{5}$ and $\cos B=\dfrac{-12}{13}$ , where $A$ and $B$ lie in first and third quadrant respectively, then $\cos (A+B)=$

1. $\dfrac{56}{65}$
2. $\dfrac{-56}{65}$
3. $\dfrac{16}{65}$
4. $\dfrac{-16}{65}$

Answer 1

In this type of question you need to use the given quantities in the question to find the value asked in the question, and for that we will first find sine and cosine of the angles that are not given in the question then we will use that quantities to find the asked value in the question.

Complete step by step answer:
Here we will use the basic trigonometric identities to solve the question, by trigonometric identities what I mean is,
Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.
As it is given in the question,
$\sin A=\dfrac{4}{5}$
Now we will use the trigonometric identity and that is:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
So after using solving above identity we get,
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$
On solving further we get,
$\Rightarrow \cos x=\pm \sqrt{1-{{\sin }^{2}}x}$
Now using above identity we derived to find cosine of the other angle we don’t know yet we get,
$\Rightarrow \cos A=\pm \sqrt{1-{{\sin }^{2}}A}$
Since we have the value of $\sin A=\dfrac{4}{5}$ , therefore putting in above equation we get,
$\Rightarrow \cos A=\pm \sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}$
Now on solving further we get,
$\Rightarrow \cos A=\pm \sqrt{1-\dfrac{16}{25}}$
$\Rightarrow \cos A=\pm \dfrac{3}{5}$
Since it can be positive or negative both so we will decide this by observing that the angle we are talking about lies in which quadrant, that is in the first and fourth quadrant cosine will always be positive and in the second and third quadrant cosine of the angle is negative.
Since it is given that,
$A$ lies in first quadrant
Therefore,
$\Rightarrow \cos A=\dfrac{3}{5}$
Till now we found the first quantity which will help in calculating the value asked in the question.
Now we have,
$\cos B=\dfrac{-12}{13}$
Now again by similar way we solved above we get,

& \Rightarrow \sin B=\pm \sqrt{1-{{\cos }^{2}}B} \\\ & \Rightarrow \sin B=\pm \sqrt{1-{{\left( \dfrac{-12}{13} \right)}^{2}}} \\\ & \Rightarrow \sin B=\pm \sqrt{1-\dfrac{144}{169}} \\\ & \Rightarrow \sin B=\pm \dfrac{5}{13} \\\ \end{aligned}$$ Now since it can be positive or negative both, but since, $$B$$ lies in third quadrant, Therefore the sign of the angle will be negative in the third quadrant. So we get, $$\Rightarrow \sin B=\dfrac{-5}{13}$$ Now, we have to find $$\cos (A+B)$$ By using formula: $$\cos (x+y)=\cos x\cos y-\sin x\sin y$$ We have, $$\begin{aligned} & \Rightarrow \cos (A+B)=\cos A\cos B-\sin A\sin B \\\ & \Rightarrow \cos (A+B)=\left( \dfrac{3}{5} \right)\left( \dfrac{-12}{13} \right)-\left( \dfrac{4}{5} \right)\left( \dfrac{-5}{13} \right) \\\ & \Rightarrow \cos (A+B)=\left( \dfrac{-36}{65} \right)-\left( \dfrac{-20}{65} \right) \\\ & \Rightarrow \cos (A+B)=\left( \dfrac{-36+20}{65} \right) \\\ & \Rightarrow \cos (A+B)=\dfrac{-16}{65} \\\ \end{aligned}$$ Hence we got the value of $$\cos (A+B)=\dfrac{-16}{65}$$ . **So, the correct answer is “Option 4”.** **Note:** Trigonometry can be used to roof a house, to make the roof inclined (in the case of single individual bungalows) and the height of the roof in buildings etc. It is used in the naval and aviation industries. It is used in cartography (creation of maps).It contributes in calculus also.