Question

cos x =5/13,sin y =-4/5 where x and y both lie on second quadrant .find the value of cos ( x + y)​

Answer 1

Answer:

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Answer 2

The value of cos(x+y) is $\frac{1}{13}$.

Given:

cos x= $\frac{5}{13}$

sin y = $\frac{ - 4}{5}$

x and y both lie on the second quadrant.

To find:

The value of cos ( x + y).

Solution:

• Trigonometric equations are the relation between the sides and angles of a right-angled triangle.
• Sine (sin) is the ratio of the perpendicular to that of the hypotenuse and cosine (cos) is the ratio of the base to that of the perpendicular.

We know that,

${ \sin }^{2} \alpha + { \cos }^{2} \alpha = 1$

Using this formula we can find sin x and cos y.

$\cos \: y = \sqrt{1- { \sin }^{2}y }$

$\cos \: y = \sqrt{1-\frac{16}{25} }$

$\cos \: y = \sqrt{ \frac{9}{25} }$

$\cos \: y = \frac{3}{5}$

Since, cos is negative in the second quadrant.

So, the value of cos y will be $\cos \: y = -\frac{3}{5}$

Similarly,

$\sin \: x = \sqrt{1 - { \cos }^{2}x }$

$\sin x= \sqrt{1 - {( \frac{5}{13}) }^{2} }$

$\sin x= \sqrt{1 - { \frac{25}{169} }}$

$\sin x= \sqrt { \frac{144}{169} }$$\sin \: x = \frac{5}{13}$

Since, sin is positive in the second quadrant.

So, the value of sin x will be $\frac{5}{13}$.

As we know,

$\cos(x + y) = \cos x \cos y - \sin x\sin y$

Put the values of sin x, sin y, cos x, and cos y in the above formula.

$\cos(x + y) = (\frac{5}{13})( \frac{ - 3}{5} ) - ( \frac{5}{13})( \frac{ - 4}{5})$

$\cos(x + y) = \frac{ - 3}{13} + \frac{4}{13}$

$\cos(x + y) = \frac{1}{13}$

Therefore, the value of cos (x+y) is $\frac{1}{13}$.

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