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Evaluate each of the following :

Problem 1 :

sin(180° - θ)

Problem 2 :

cos(180° - θ)

Problem 3 :

tan(180° - θ)

Problem 4 :

csc(180° - θ)

Problem 5 :

sec(180° - θ)

Problem 6 :

cot(180° - θ)

Problem 7 :

sin(180° + θ)

Problem 8 :

cos(180° + θ)

Problem 9 :

tan(180° + θ)

Problem 10 :

csc(180° + θ)

Problem 11 :

sec(180° + θ)

Problem 12 :

cot(180° + θ)

Problem 13 :

Find two angles that satisfy the equation.

sin (2x - 4) = cos (3x + 9)

Answers

1. Answer :

sin(180° - θ)

To evaluate sin(180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "sin" will not be changed as "cos".

(iii) In the II nd quadrant, the sign of "sin" is positive.

Considering the above points, we have

sin(180° - θ) = sinθ

2. Answer :

cos(180° - θ)

To evaluate cos(180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "cos" will not be changed as "sin".

(iii) In the II nd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos(180° - θ) = -cosθ

3. Answer :

tan(180° - θ)

To evaluate tan(180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "tan" will not be changed as "cot".

(iii) In the II nd quadrant, the sign of "tan" is negative.

Considering the above points, we have

tan(180° - θ) = -tanθ

4. Answer :

csc(180° - θ)

To evaluate csc(180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "csc" will not be changed as "sec".

(iii) In the II nd quadrant, the sign of "csc" is positive.

Considering the above points, we have

csc(180° - θ) = cscθ

5. Answer :

sec(180° - θ)

To evaluate sec(180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "sec" will not be changed as "csc".

(iii) In the II nd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec(180° - θ) = -secθ

6. Answer :

cot(180° - θ)

To evaluate cot(180° - θ), we have to consider the following important points.

(i) (180° - θ) will fall in the II nd quadrant.

(ii) When we have 180°, "cot" will not be changed as "tan".

(iii) In the II nd quadrant, the sign of "cot" is negative.

Considering the above points, we have

cot(180° - θ) = -cotθ

7. Answer :

sin(180° + θ)

To evaluate sin(180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "sin" will not be changed as "cos".

(iii) In the III rd quadrant, the sign of "sin" is negative.

Considering the above points, we have

sin(180° + θ) = -sinθ

8. Answer :

cos(180° + θ)

To evaluate cos(180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "cos" will not be changed as "sin".

(iii) In the III rd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos(180° + θ) = -cosθ

9. Answer :

tan(180° + θ)

To evaluate tan(180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "tan" will not be changed as "cot".

(iii) In the III rd quadrant, the sign of "tan" is positive.

Considering the above points, we have

tan(180° + θ) = tanθ

10. Answer :

csc(180° + θ)

To evaluate csc(180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "csc" will not be changed as "sec".

(iii) In the III rd quadrant, the sign of "csc" is negative.

Considering the above points, we have

csc(180° + θ) = -cscθ

11. Answer :

sec(180° + θ)

To evaluate sec(180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "sec" will not be changed as "csc".

(iii) In the III rd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec(180° + θ) = -secθ

12. Answer :

cot(180° + θ)

To evaluate cot(180° + θ), we have to consider the following important points.

(i) (180° + θ) will fall in the III rd quadrant.

(ii) When we have 180°, "cot" will not be changed as "tan".

(iii) In the III rd quadrant, the sign of "cot" is positive.

Considering the above points, we have

cot(180° + θ) = cotθ

13. Answer :

sin (2x - 4) = cos (3x + 9)

sin x = cos (90 - x)

cos (90 - (2x - 4)) = cos (3x + 9)

90 - (2x - 4) = 3x + 9

90 - 2x + 4 = 3x + 9

94 - 2x = 3x + 9

-2x - 3x = 9 - 94

-5x = -85

x = 85/5

x = 17

Applying x = 17 in (2x - 4), we get

= 2(17) - 4

= 34 - 4

= 30

Applying x = 17 in (3x + 9), we get

= 3(17) + 9

= 51 + 9

= 60

So, the required angles are 30 degree and 60 degree.

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