Question

If a and b are two angles in standard position in Quadrant I, find cos(a+b) for the given function values. sin a=15/17and cos b=3/5
1) -84/85

2) -36/85

3) 36/85

4) 84/85

Answer 1

The value of $cos(a+b)$ for the angles $a$ and $b$ in standard position in the first quadrant is $-\frac{36}{85}$

We need to find the value of $cos(a+b)$. To proceed, we need to use the compound angle formula

Cosine of a sum of two angles

The cosine of the sum of two angles $a$ and $b$ is given below

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

We are given

$sin(a)=\dfrac{15}{17}\\\\cos(b)=\dfrac{3}{5}$

We need to find $sin(b)$ and $cos(a)$, using the identity

$sin^2(\theta)+cos^2(\theta)=1$

Find sin(b)

To find $sin(b)$, note that

$sin^2(b)+cos^2(b)=1\\\\\implies sin(b)=\sqrt{1-cos^2(b)}$

substituting $\frac{3}{5}$ for $cos(b)$ in the identity, we get

$sin(b)=\sqrt{1-cos^2(b)}\\\\=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}$

Find cos(a)

To find $cos(a)$, note that

$sin^2(a)+cos^2(a)=1\\\\\implies cos(a)=\sqrt{1-sin^2(a)}$

substituting $\frac{15}{17}$ for $sin(a)$ in the identity, we get

$cos(a)=\sqrt{1-sin^2(a)}\\\\=\sqrt{1-\left(\dfrac{15}{17}\right)^2}=\dfrac{8}{17}$

Find the value of cos(a+b)

We can now make use of the formula

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

to find $cos(a+b)$.

$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)\\\\=\dfrac{8}{17}\cdot\dfrac{3}{5}-\dfrac{15}{17}\cdot\dfrac{4}{5}=-\dfrac{36}{85}$

Learn more about sine and cosine of compound angles here brainly.com/question/24305408