Trigonometrical Ratios of the Sum and Difference of Two Angles

Trigonometrical Ratios of the Sum and Difference of Two Angles

We will start with a list of the formulae regarding Trigonometrical Ratios of the Sum and Difference of Two Angles. Skip to the examples here. Alternatively, you may choose to start over with the Measurement of Angles.

Addition & Subtraction Theorems

\[
\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)
\]\[
\sin(A – B) = \sin(A)\cos(B) – \cos(A)\sin(B)
\]\[
\cos(A + B) = \cos(A)\cos(B) – \sin(A)\sin(B)
\]\[
\cos(A – B) = \cos(A)\cos(B) + \sin(A)\sin(B)
\]

1.

\[
\sin(C) + \sin(D) = 2\sin\mkern-5mu\left( \frac{C + D}{2} \right)\cos\mkern-5mu\left( \frac{C – D}{2} \right)
\]\[
\sin(C) – \sin(D) = 2\cos\mkern-5mu\left( \frac{C + D}{2} \right)\sin\mkern-5mu\left( \frac{C – D}{2} \right)
\]\[
\cos(C) + \cos(D) = 2\cos\mkern-5mu\left( \frac{C + D}{2} \right)\cos\mkern-5mu\left( \frac{C – D}{2} \right)
\]\[
\cos(C) – \cos(D) = 2\sin\mkern-5mu\left( \frac{C + D}{2} \right)\sin\mkern-5mu\left( \frac{D – C}{2} \right)
\]

2.

\[
2\sin(A)\cos(B) = \sin(A + B) + \sin(A – B)
\]\[
2\cos(A)\sin(B) = \sin(A + B) – \sin(A – B)
\]\[
2\cos(A)\cos(B) = \cos(A + B) + \cos(A – B)
\]\[
2\sin(A)\sin(B) = \cos(A – B) – \cos(A + B)
\]

3.

\[
\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 – \tan(A)\tan(B)}
\]\[
\tan(A – B) = \frac{\tan(A) – \tan(B)}{1 + \tan(A)\tan(B)}
\]

Study Inverse Trigonometric Functions

SL Loney – Examples 13

Prove that:

#4. \(\cos(45{°} – A)\cos(45{°} – B) – \sin(45{°} – A)\sin(45{°} – B) = \sin(A + B)\)

Solution: LHS \(= \cos(45{°} – A + 45{°} – B) = \cos(90{°} – (A + B)) = \sin(A + B) = \) RHS

#5. \(\sin(45{°} + A)\cos(45{°} – B) + \cos(45{°} + A)\sin(45{°} – A) = \cos(A – B)\)

#6. \(\frac{\sin(A – B)}{\cos(A)\cos(B)} + \frac{\sin(B – C)}{\cos(B)\cos(C)} + \frac{\sin(C – A)}{\cos(C)\cos(A)} = 0 \)

Solution: Here, \[ LHS = \frac{\sin(A)\cos(B) – \cos(A)\sin(B)}{\cos(A)\cos(B)} + \frac{\sin(B)\cos(C) – \sin(C)\cos(B)}{\cos(B)\cos(C)} \]\[+ \frac{\sin(C)\cos(A) – \cos(C)\sin(A)}{\cos(A)\cos(C)}\]

\[= \frac{\sin(A)\cos(B)}{\cos(A)\cos(B)} – \frac{\cos(A)\sin(B)}{\cos(A)\cos(B)} + \frac{\sin(B)\cos(C)}{\cos(B)\cos(C)} – \frac{\sin(C)\cos(B)}{\cos(B)\cos(C)} \]\[+\frac{\sin(C)\cos(A)}{\cos(A)\cos(C)} – \frac{\cos(C)\sin(A)}{\cos(A)\cos(C)}\] \[ = 0 = RHS \]

#7. \(\sin105{°} + \cos105{°} = \cos45{°}\)

Solution: \[
LHS = \sin(60{°} + 45{°}) + \cos(60{°} + 45{°})
\]\[
= \sin60{°}\cos45{°} + \cos60{°}\sin45{°} + \cos60{°}\cos45{°} – \sin60{°}\sin45{°}
\]\[
= \cos45{°} = RHS
\]

SL Loney – Examples 14

Prove that:

#1. \(\frac{\sin(7\theta) – \sin(5\theta)}{\cos(7\theta) + \cos(5\theta)} = \tan(\theta)\)

#3. \(\frac{\sin(A) + \sin(3A)}{\cos(A) + \sin(3A)} = \tan(2A)\)

Solution: LHS \[
= \frac{2\sin\mkern-5mu\left( \frac{3A + A}{2} \right)\cos\mkern-5mu\left( \frac{3A – A}{2} \right)}{2\cos\mkern-5mu\left( \frac{3A + A}{2} \right)\cos\mkern-5mu\left( \frac{3A – A}{2} \right)}
\]\[
= \frac{2\sin(2A)\cos(A)}{2\cos(2A)\cos(A)} = \tan2A
\]

#6. \(\frac{\sin(2A) + \sin(2B)}{\sin(2A) – \sin(2B)} = \frac{\tan(A + B)}{\tan(A – B)}\)

Solution: \[
= \frac{2\sin(A + B)\cos(A – B)}{2\cos(A + B)\sin(A – B)}
\]\[
= \tan(A + B)\cot(A – B)
\]\[
= \frac{\tan(A + B)}{\tan(A – B)}
\]
#9. \(\frac{\cos(2B) – \cos(2A)}{\sin(2A) + \sin(2B)} = \tan(A – B)\)

Solution: LHS \(=\)
\[
\frac{2\sin\mkern-5mu\left( \frac{2B + 2A}{2} \right)\sin\mkern-5mu\left( \frac{2A – 2B}{2} \right)}{2\sin\mkern-5mu\left( \frac{2A + 2B}{2} \right)\cos\mkern-5mu\left( \frac{2A – 2B}{2} \right)} \]\[= \frac{2\sin(A + B)\sin(A – B)}{2\sin(A + B)\cos(A – B)} \]\[= \tan(A – B) = RHS
\]

#10. \(\cos(A + B) + \sin(A – B) = 2\sin(45{°} + A)\cos(45{°} + B)\)

Hint: Use \( \sin(90{°} + \theta) = \cos(\theta) \) and \[
\sin(C) + \sin(D) = 2\sin\mkern-5mu\left( \frac{C + D}{2} \right)\cos\mkern-5mu\left( \frac{C – D}{2} \right)
\]

#15. \(\frac{\sin(A) + \sin(3A) + \sin(5A) + \sin(7A)}{\cos(A) + \cos(3A)\cos(5A) + \cos(7A)} = \tan(4A)\)

Solution: LHS \(=\)\[
\frac{2\sin(2A)\cos(A) + 2\sin(6A)\cos(A)}{2\cos(2A)\cos(A) + 2\cos(6A)\cos(A)}
\]\[
= \frac{\sin(2A) + \sin(6A)}{\cos(2A) + \cos(6A)} = \frac{2\sin(4A)\cos(2A)}{2\cos(4A)\cos(2A)}
\]\[
= \tan(4A)
\]

#20. Prove that \[
\frac{\sin(A) + \sin(B)}{\sin(A) – \sin(B)} = \tan\mkern-5mu\left( \frac{A + B}{2} \right)\cot\mkern-5mu\left( \frac{A – B}{2} \right)
\]

#22. Prove that \[
\frac{\sin(A) + \sin(B)}{\cos(A) + \cos(B)} = \tan\mkern-5mu\left( \frac{A + B}{2} \right)
\]

SL Loney – Examples 15

#8. \(\{\sin(3A) + \sin(A)\}\sin(A) + \{\cos(3A) – \cos(A)\}\cos(A) = 0\)

Solution: \[
= (2\sin(2A)\cos(A))\sin(A) + (2\sin(2A)\sin(-A))\cos(A)
\]\[
= 2\sin(2A)\cos(A)\sin(A) – 2\sin(2A)\sin(A)\cos(A) = 0 = RHS
\]

#14. \(\sin(45{°} + A)\sin(45{°} – A) = \frac{1}{2}\cos(2A)\)

Solution: LHS \(=\)

\[
\frac{2\sin(45{°} + A)\sin(45{°} – A)}{2}
\]

\[
= \frac{\cos(2A) – \cos(90{°})}{2} = \frac{1}{2}\cos(2A) = RHS
\]

SL Loney – Examples 16

#3. If \(\tan(A) = \frac{n}{n + 1}\) and \(\tan(B) = \frac{1}{2n + 1}\), find \(\tan(A + B).\)
Answer: 1

#7. Prove that \[
1 + \tan(A)\tan\mkern-5mu\left( \frac{A}{2} \right) = \tan(A)\cot\mkern-5mu\left( \frac{A}{2} \right) – 1 = \sec(A)
\]
Solution: Taking first, \[
1 + \tan(A)\tan\mkern-5mu\left( \frac{A}{2} \right)
\]\[
= 1 + \frac{\sin(A)\sin\mkern-5mu\left( \frac{A}{2} \right)}{\cos(A)\cos\mkern-5mu\left( \frac{A}{2} \right)}
\]\[
= \frac{\cos(A)\cos\mkern-5mu\left( \frac{A}{2} \right) + \sin(A)\sin\mkern-5mu\left( \frac{A}{2} \right)}{\cos(A)\cos\mkern-5mu\left( \frac{A}{2} \right)}
\]\[
= \frac{\cos\mkern-5mu\left( A – \frac{A}{2} \right)}{\cos(A)\cos\mkern-5mu\left( \frac{A}{2} \right)}
\]\[
= \frac{\cos\mkern-5mu\left( \frac{A}{2} \right)}{\cos(A)\cos\mkern-5mu\left( \frac{A}{2} \right)} = \sec(A) = RHS
\]

Next, take \[
\cot\mkern-5mu\left( \frac{A}{2} \right)\tan(A) – 1
\]\[
= \frac{\cos\mkern-5mu\left( \frac{A}{2} \right)\sin(A)}{\sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)} – 1
\]\[
= \frac{\cos\mkern-5mu\left( \frac{A}{2} \right)\sin(A) – \sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)}{\sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)}
\]\[
= \frac{\sin(A)\cos\mkern-5mu\left( \frac{A}{2} \right) – \sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)}{\sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)}
\]\[
= \frac{\sin\mkern-5mu\left( A – \frac{A}{2} \right)}{\sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)} = \frac{\sin\mkern-5mu\left( \frac{A}{2} \right)}{\sin\mkern-5mu\left( \frac{A}{2} \right)\cos(A)} = \sec(A) = RHS
\]

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