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=3/5and cos b=12/371) 153/1852) 57/1853) -57/1854) -153/185

Answer 1

The Solution:

Given that a and b are two standard angles in the first quadrant.

We are required to find cos(a+b) if

$\begin{gathered} \sin a=\frac{3}{5},\cos b=\frac{12}{37} \\ \end{gathered}$

By the formula of compound angles,

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

Step 1:

We need to find the value of:

$\cos a\text{ and }\sin b$

Given that:

$\begin{gathered} \sin a=\frac{3}{5}\text{ , and angle a is in the first quadrant, that is,} \\ 0^oBy Trigonometry we have:

By the Pythagorean Theorem, we can find x as below:

[tex]\begin{gathered} 5^2=3^2+x^2 \\ 25=9+x^2 \\ 25-9=x^2 \\ 16=x^2 \end{gathered}$

Taking the square root of both sides, we get

$\begin{gathered} \sqrt[]{16}=\sqrt[]{x^2} \\ \pm4=x \\ x=4 \end{gathered}$

So, we can get the value of:

$\begin{gathered} \cos a=\frac{adj}{hyp}=\frac{4}{5} \\ \\ \cos a=\frac{4}{5} \end{gathered}$

Similarly, we shall find sinb :

$\text{ If }\cos b=\frac{12}{37},0^o

By the Pythagorean Theorem, we can find y as below:

[tex]\begin{gathered} y^2+12^2=37^2 \\ y^2+144=1369 \\ y^2=1369-144 \\ y^2=1225 \end{gathered}$

Taking the square root of both sides, we get

$\begin{gathered} \sqrt[]{y^2}=\text{ }\sqrt[]{1225} \\ \\ y=35 \end{gathered}$

So, we can get the value of:

$\begin{gathered} \sin b=\frac{opp}{hyp}=\frac{35}{37} \\ \\ \sin b=\frac{35}{37} \end{gathered}$

Substituting these values in the formula below:

$\begin{gathered} \cos (a+b)=\cos a\cos b-\sin a\sin b \\ \\ \cos (a+b)=(\frac{4}{5}\times\frac{12}{37})-(\frac{3}{5}\times\frac{35}{37}) \end{gathered}$$\begin{gathered} \cos (a+b)=\frac{48}{185}-\frac{105}{185}=\frac{-57}{185} \\ \\ \cos (a+b)=-\frac{57}{185} \end{gathered}$

Therefore, the correct answer is [option 3]