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) ?

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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 :)

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Answer 1 · 2018-04-13T17:14:18

See below.

Explanation:

We are given:

$sin A = - \frac{3}{5}$ $sinA=-3/5$

Squaring both sides:

${sin}^{2} A = \frac{9}{25}$ $sin^2A=9/25$

Identity:

${sin}^{2} x + {cos}^{2} x = 1$ $color(red)bb(sin^2x+cos^2x=1)$

i.e.

${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$

Substituting this:

$1 - {cos}^{2} A = \frac{9}{25}$ $1-cos^2A=9/25$

${cos}^{2} A = \frac{16}{25}$ $cos^2A=16/25$

Taking roots:

$cos A = \pm \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}$ $cosA=+-(sqrt(16))/sqrt(25)=4/5$

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

$cos A = \frac{4}{5}$ $color(blue)(cosA=4/5$

For $cos B$ $cosB$ we use the same idea:

$sin B = - \frac{1}{3}$ $sinB=-1/3$

${sin}^{2} B = \frac{1}{9}$ $sin^2B=1/9$

$1 - {cos}^{2} B = \frac{1}{9}$ $1-cos^2B=1/9$

${cos}^{2} B = \frac{8}{9}$ $cos^2B=8/9$

Taking roots:

$cos B = \pm \frac{2 \sqrt{2}}{3}$ $cosB=+-(2sqrt(2))/3$

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

$cos B = \frac{2 \sqrt{2}}{3}$ $color(blue)(cosB=(2sqrt(2))/3)$

Using identity:

$tan x = \frac{sin x}{cos x}$ $color(red)bb(tanx=sinx/cosx)$

Find:

$tan A and tan B$ $tanA and tanB$

$tan A = \frac{sin A}{cos A} = \frac{- \frac{3}{5}}{\frac{4}{5}} = - \frac{3}{4}$ $tanA=sinA/cosA=(-3/5)/(4/5)=-3/4$

$tan B = \frac{sin B}{cos B} = \frac{- \frac{1}{3}}{\frac{2 \sqrt{2}}{3}} = - \frac{1}{2 \sqrt{2}} = - \frac{\sqrt{2}}{4}$ $tanB=sinB/cosB=(-1/3)/((2sqrt(2))/3)=-1/(2sqrt(2))=-sqrt(2)/4$

Identities:

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

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

$:.$

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

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

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

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

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

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

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

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

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

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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 :)

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

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 :)

1 Answer

Explanation:

Related questions

Impact of this question

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

Thanks :)