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Chapter 7: Trigonometric Ratios of Compound Angles – Exercise 7.1

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.1

We have,

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.1

Now,

sin(A + B) = sin A cos B + cos A sin B

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.1(i)

We have, 

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.1(ii)

Now,

cos(A + B) = cos A cos B - sin A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.1(iii)

We have,

trigonometric-ratios-of-compound-angles–exercise-7.1–q.1(iv)

Now,

Sin (A - B) = sin A cos B - cos A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.1(v)

We have,

trigonometric-ratios-of-compound-angles–exercise-7.1–q.1(vi)

Now

cos(A - B) = cos A cos B + sin A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.1(vii)

 

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.2

We have,

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.2

Now,

(i) sin(A + B) = sin A cos B + cos A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.2(i)

(ii) cos(A + B) = cos A cos B - sin A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.2(ii)

We have,

trigonometric-ratios-of-compound-angles–exercise-7.1–q.2(iii)

Now,

sin(A+B) = sin A cos B + cos A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.2(iv)

 

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.3

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.3

Now,

(i) sin(A + B) = sin A cos B + cos A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.3(i)

(ii) cos(A+B) = cos A cos B - sin A sin B

trigonometric-ratios-of-compound-angles–exercise-7.1–q.3(ii)

 

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.4

We have,

Trigonometric Ratios of Compound Angles – Exercise 7.1 – Q.4

Now,

trigonometric-ratios-of-compound-angles–exercise-7.1–q.4(i)


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