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=(3)/(5),\cos b=(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:

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=(adj)/(hyp)=(4)/(5) \\ \\ \cos a=(4)/(5) \end{gathered}`

Similarly, we shall find sinb :

`\text{ If }\cos b=(12)/(37),0^o<p>By the Pythagorean Theorem, we can find y as below:</p>[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=(opp)/(hyp)=(35)/(37) \\ \\ \sin b=(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)=((4)/(5)*(12)/(37))-((3)/(5)*(35)/(37)) \end{gathered}`
`\begin{gathered} \cos (a+b)=(48)/(185)-(105)/(185)=(-57)/(185) \\ \\ \cos (a+b)=-(57)/(185) \end{gathered}`

Therefore, the correct answer is [option 3]