# Given: `csc theta = −17/15` , where 270° ≤ θ ≤ 360° and `cot beta = −3/4` , where 90° ≤ β ≤ 180°.Find the exact value of sin(θ + β).Math Grade 11 Given: csc θ =...

Given:

`csc theta = −17/15` , where 270° ≤ θ ≤ 360° and

`cot beta = −3/4` , where 90° ≤ β ≤ 180°.

Find the exact value of sin(θ + β).

Math Grade 11

Given:

csc θ = −(17/15), where 270° ≤ θ ≤ 360° and

cot β = −(3/4) where 90° ≤ β ≤ 180°.

Find the exact value of sin(θ + β). Show all the work.

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### 2 Answers

Using the identity of sum of two angles,

`sin(theta+beta)=sinthetacosbeta + costhetasinbeta`

Next, determine the values of sine and cosine base on the given trigonometric functions.

From `csc theta =-17/15` , where `270^olt=thetalt360^o` , solve for `sin theta` and `cos theta` .

Note that `csc theta= (hypoten u se)/(opposite)` .

So, hypotenuse=17 and opposite=15.

Then, apply the pythagorean formula to determine the length of the side adjacent to theta.

`adjacent = sqrt ((hypoten us e)^2-(opposite)^2) = sqrt(17^2-15^2) =8`

And use the formula for sine and cosine. Note that `theta` lies at the fourth quadrant, so sign of cosine is positve and negative for sine.

`sin theta =-(opposite)/(hypoten u se)=-15/17`

`cos theta = -(adjacent)/(hyponten u se)=8/17`

Next, determine `sin beta` and `cos beta` using the given `cot beta=-3/4` , where `90^olt=betalt=180^o` .

The formula for cotangent is `cot beta=(adjacent)/(opposite)` .

Hence, adjacent=3 and opposite=4.

Then, apply the pythagorean formula to determine the length of the hypotenuse.

`hypoten u se=sqrt((opposite)^2+(adjacent)^2)=sqrt(4^2+3^2)=5`

Substitute the values of adjacent,opposite and hypotenuse to the formula of sine and cosine. And since `beta` lies on the second quadrant, the sign of cosine is negative. While the sine function is positive.

`sin beta=(opposite)/(hypoten u se)=4/5`

`cos beta=-(adjacent)/(hypoten u se)=-3/5`

Then, substitute the values of `sin theta` , `cos theta` , `sin beta` and `cos beta` to the identity of sum of two angles.

`sin(theta+beta)=sinthetacosbeta + costhetasinbeta`

`sin(theta+beta)=(-15/17)*(-3/5)+(-8/17)*(4/5)`

`sin(theta+beta)=45/85+32/85 `

`sin(theta+beta)=77/85`

**Hence, `sin (theta + beta)=77/85` .**

You should expand `sin(theta + beta)` to check what you need to find, based on the information provided by the problem such that:

`sin(theta + beta) = sin theta*cos beta + sin beta* cos theta`

Since the problem provides the value of `csc theta,` you may find `sin theta` based on the following formula such that:

`csc theta = 1/sin theta => sin theta = 1/ csc theta`

`sin theta = 1/(-17/15) => sin theta = -15/17`

Since, `theta` is in quadrant 4, then the value of `sin theta ` needs to be negative, thus `sin theta = -15/17` .

You may find `cos theta` using the fundamental formula of trigonometry such that:

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

`cos^2 theta = 1 - (15/17)^2 => cos^2 theta = (17-15)(17+15)/(17^2)`

`cos^2 theta = 2*32/(17^2) => cos^2 theta = 2^6/17^2`

Since, `theta` is in quadrant 4, then the value of `cos theta` needs to be positive, thus `cos theta = sqrt(2^6/17^2) = 8/17` .

You may find `sin beta` using the following formula:

`1 + cot^2 beta = 1/(sin^2 beta)`

`1 + 9/16 = 1/(sin^2 beta) => 25/16 = 1/(sin^2 beta) => sin^2 beta = 16/25 `

Since `beta` is in quadrant 2, then `sin beta` is positive such that:

`sin beta = sqrt(16/25) = 4/5`

You may find `cos theta` using the fundamental formula of trigonometry such that:

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

`cos^2 beta = 1 - 16/25 => cos^2 beta = 9/25`

Since `beta` is in quadrant 2, then `cos beta` is negative such that:

`cos beta = -sqrt(9/25) = -3/5`

Substituting the found values in formula `sin(theta + beta) = sin theta*cos beta + sin beta* cos theta` yields:

`sin(theta + beta) = (-15/17)*(-3/5)+ 4/5* 8/17`

`sin(theta + beta) = (45 + 32)/85 => sin(theta + beta) = 77/85`

**Hence, evaluating `sin(theta + beta)` using trigonometric identities above yields `sin(theta + beta) = 77/85` .**