Example.1) Find the value (without using table) –
(i) (cos 0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
Ans.) Substituting the value of given T-Ratios such as –
cos 0⁰ = 1, sin 45⁰ = 1/√2, sin 30⁰ = 1/2, sin 90⁰ = 1, cos 45⁰ = 1/√2, cos 60⁰ = 1/2.
Now, we get –
(cos 0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
1 1 1 1
= (1 + ------ + ------) (1 - ------- + ------)
√2 2 √2 2
2 + √2 2 - √2
= (1 + ----------) {1 – (-----------)
2√2 2√2
2√2 + 2 + √2 2√2 – 2 + √2
= (----------------) (---------------)
2√2 2√2
8 + 4√2 + 4 - 4√2 – 4 - 2√2 + 4 + 2√2 + 2
= (------------------------------------------)
8
14 7
= (-------) = ------- (Ans.)
8 4
Example.2) If, A = 60⁰, verify that –
(i) sin² A + cos² A = 1
Ans.) (sin A)² + (cos A)²
Replacing A = 60⁰, we get –
(sin 60⁰)² + (cos 60²)²
√3 1 3 1
= (------)² + (------)² = (------ + ------)
2 2 4 4
4
= (-----) = 1 (Ans.)
4
(ii) cosec² A + cos² A = 19/12
Ans.) cosec² A + cos² A = (cosec A)² + (cos A)²
Replacing A = 60⁰, we get –
(cosec 60⁰)² + (cos 60⁰)²
Now, replacing the value of cosec 60⁰ & sec 60⁰ and
2 1 4 1
= (-------)² + (------)² = (------- + ------)
√3 2 3 4
16 + 3 19
= ----------- = ------- (Ans.)
12 12
Example.3) Taking A = 30⁰, verify that –
2 tan A
(i) tan 2A = --------------
(1 - tan² A)
Ans.) as per the given condition A = 30⁰
Or, 2A = 60⁰
2 tan A
tan 2A = --------------
(1 - tan² A)
=> LHS = RHS
Now, LHS = tan 2A = tan 60⁰ = √3
2 tan A
And, RHS = --------------
(1 - tan² A)
Substituting the value of A = 30⁰, and we get –
2 tan 30⁰
RHS = ----------------
(1 - tan² 30⁰)
1
Substitute the value of tan 30⁰ = -------, and we get -
√3
2 tan 30⁰ 2 X (1/√3) 2/√3
-------------- = --------------- = -----------
(1 - tan² 30⁰) {1 – (1/√3)²} (1 – 1/3)
2/√3
= ----------
2/3
2 2
= ------- ÷ -------
√3 3
2 3 3
= ------- X ------ = -------
√3 2 √3
√3 X √3
= ----------- = √3
√3
(1 – cos 2A)
(ii) sin A = √--------------
2
Ans.) as per the given condition A = 30⁰
Or, 2A = 60⁰
(1 – cos 2A)
sin A = √---------------
2
=> LHS = RHS
1
So, LHS = sin A = sin 30⁰ = -------
2
(1 – cos A) (1 – cos 30⁰)
Now, RHS = √------------- = √---------------
2 2
1
(1 - -----)
2
= √----------------------
2
1/2 1
= √------- = √------ ÷ 2
2 2
1 1
= √(----- X ------)
2 2
1 1
= √ ------ = ------
4 2
So, LHS = RHS (Proved)
Example.4) Taking A = 30⁰, verify that –
sin 3A = (3 sin A – 4 sin³ A)
Ans.) As per the given condition A = 30⁰
Or, 3A = 90⁰
sin 3A = (3 sin A – 4 sin³ A)
=> LHS = RHS
LHS = sin 3A= sin 90⁰ = 1
And, RHS = (3 sin A – 4 sin³ A)
= {3 sin A – 4 (sin A)³}
Now, we will substitute the value of A = 30⁰, and we get –
= {3 sin 30⁰ – 4 (sin 30⁰)³}
Now, we will substitute the value of sin 30⁰ = 1/2, and we get –
1 1
= {(3 X ------) – (4 X ------)}
2 2³
3 1 3 1
= {------ - (4 X ------)} = (------- - -------)
2 8 2 2
3 – 1 2
= (-------) = ------- = 1 (Proved)
2 2
Example.5) If A = 60⁰, & B = 30⁰, prove that-
Sin (A – B) = sin A cos B – cos A sin B
Ans.) as per the given condition A = 60⁰, & B = 30⁰
sin (A – B) = sin A cos B – cos A sin B
=> LHS = RHS
LHS = sin (A – B)
= sin (60 – 30)⁰ = sin 30⁰ = 1/2
Now, RHS = sin A cos B – cos A sin B
= sin 60⁰ cos 30⁰ - cos 60⁰ sin 30⁰
√3 √3 1 1
= (------- X -------) – (------X ------)
2 2 2 2
3 1 (3 – 1)
= (------- - -------) = ----------
4 4 4
2 1
= ------- = --------
4 2
So, LHS = RHS (Proved)