Question

If A is an angle in QIV such that sin A = − 3 5 and B is an angle in QIV such that sin B = − 1 3 use identities to find COS A COS B TAN(A+B) TAN(A-B) ?

Let me clarify that it is -3/5 and -1/3

Thanks :)

Answer 1

See below.

Explanation:

We are given:

`sinA=-3/5`

Squaring both sides:

`sin^2A=9/25`

Identity:

`color(red)bb(sin^2x+cos^2x=1)`

i.e.

`sin^2x=1-cos^2x`

Substituting this:

`1-cos^2A=9/25`

`cos^2A=16/25`

Taking roots:

`cosA=+-(sqrt(16))/sqrt(25)=4/5`

Since we are in the IV quadrant we expect the cosine to be positive:

`color(blue)(cosA=4/5`

For `cosB` we use the same idea:

`sinB=-1/3`

`sin^2B=1/9`

`1-cos^2B=1/9`

`cos^2B=8/9`

Taking roots:

`cosB=+-(2sqrt(2))/3`

Since we are in the IV quadrant we expect the cosine to be positive:

`color(blue)(cosB=(2sqrt(2))/3)`

Using identity:

`color(red)bb(tanx=sinx/cosx)`

Find:

`tanA and tanB`

`tanA=sinA/cosA=(-3/5)/(4/5)=-3/4`

`tanB=sinB/cosB=(-1/3)/((2sqrt(2))/3)=-1/(2sqrt(2))=-sqrt(2)/4`

Identities:

`color(red)bb(tan(A+B)=(tanA+tanB)/(1-tanAtanB))`

`color(red)bb(tan(A-B)=(tanA-tanB)/(1+tanAtanB))`

`:.`

`tan(A+B)=((-3/4)+(-sqrt(2)/4))/(1-(-3/4)(-sqrt(2)/4))=((-3-sqrt(2))/4)/((16-3sqrt(2))/16)->`

`=(-12-4sqrt(2))/(16-3sqrt(2))=color(blue)(-((12+4sqrt(2)))/(16-3sqrt(2)))`

`tan(A-B)=((-3/4)-(-sqrt(2)/4))/(1+(-3/4)(-sqrt(2)/4))=((-3+sqrt(2))/4)/((16+3sqrt(2))/16)->`

`=(-12+4sqrt(2))/(16+3sqrt(2))=color(blue)(-((12-4sqrt(2)))/(16+3sqrt(2))`

.....................................................................................................................................

`color(blue)(cosA=4/5`

`color(blue)(cosB=(2sqrt(2))/3)`

`color(blue)(tan(A+B)=-((12+4sqrt(2)))/(16-3sqrt(2)))`

`color(blue)(tan(A-B)=-((12-4sqrt(2)))/(16+3sqrt(2))`