Conversion of Trigonometric Ratios

Conversion of Trigonometric Ratios

Conversion of Trigonometric Ratios

There are six trigonometric ratios. If one of them is known, the remaining five can be obtained by the conversion of trigonometric ratios. If we know any one of the trigonometric ratio, then the rest can be converted in terms of that known ratio. This can be done by using either of the following two methods:

(a) Using basic trigonometric relations

(b) Using Pythagoras theorem

 

Lets study the following worked out examples:


Example 1: Express all the trigonometric ratios of angle A in terms of sinA. Solution: Here, 	sinA = sinA 	cosecA = 1/sinA cosA = √(1-〖sin〗^2 A) secA = 1/cosA = 1/√(1-〖sin〗^2 A) tanA = sinA/cosA = sinA/√(1-〖sin〗^2 A) cotA = cosA/sinA = √(1-〖sin〗^2 A)/sinA Hence, all the trigonometric ratios are expressed in terms of sinA.


Example 2: If Sinθ = 12/13, find cosθ and tanθ. Solution: Here, 	Sinθ = 12/13 = p/h ∴	p = 12 and h = 13 ∴	b = √(h^2-p^2 ) = √(〖13〗^2-〖12〗^2 ) = √(169-144) = √25 = 5 ∴	cosθ = b/h = 5/13  And,      tanθ = p/b = 12/5


Example 3: If cosθ = m/√(m^2+n^2 ) , then prove that: m sin⁡〖θ=n cos⁡θ 〗. Solution: Here,  cosθ = m/√(m^2+n^2 ) = b/h ∴	b = m and h = √(m^2+n^2 ) ∴ p = √(h^2-b^2 ) = √(〖(√(m^2+n^2 ))〗^2-m^2 ) = √(m^2+n^2-m^2 ) = √(n^2 ) = n Now, 	tanθ = p/b i.e.  sin⁡θ/cos⁡θ  = n/m or,	 m sin⁡〖θ=n cos⁡θ 〗.  Proved.


Example 4: If 3Sinθ + 4cosθ = 5, show that cosθ = 4/5.

Solution: Here,

          3sinθ + 4cosθ = 5

or,     5 – 4cosθ = 3sinθ

          Squaring both sides, we get

          25 – 40cosθ + 16cos2θ = 9sin2θ

or,     16cos2θ – 40cosθ + 25 = 9(1 – cos2θ)

or,     16cos2θ – 40cosθ + 25 = 9 – 9cos2θ

or,     25cos2θ – 40cosθ + 16 = 0

or,     (5cosθ – 4)2 = 0

or,     5cosθ – 4 = 0

or,     5cosθ = 4

or,     cosθ = 4/5. Shown.


Example 5: If tanA = 1/2 and tanB = 1/3 , then show that cosA.sinB + sinA.cosB = 1/√2. Solution: Here, 	tanA = 1/2 = p/b ∴	p = 1 and b = 2 ∴	h = √(p^2+b^2 ) = √(1^2+2^2 ) = √(1+4)  = √5 ∴	sinA = p/h = 1/√5  and cosA =  b/h = 2/√5 Again,	tanB = 1/3 = p/b ∴	p = 1 and b = 3 ∴	h = √(p^2+b^2 ) = √(1^2+3^2 ) = √(1+9) = √10 ∴	sinB = p/h = 1/√10  and cosB =  b/h = 3/√10 Now, cosA.sinB + sinA.cosB = 2/√5  .  1/√10+1/√5  .  3/√10 = 2/(√5 √10)  + 3/(√5 √10) = (2+3)/(√5 √10) = 5/(√5 √(5×2)) = (√5  .√5)/(√5 √5.√2) = 1/√2  proved.


Example 6: If sinα = cosβ = √3/2 , then show that (tan⁡α  -〖 tan〗⁡β)/(1 +〖 tan〗⁡α  tan⁡β )=1/√3 . Solution: Here, 	sinα = √3/2  ∴	cosα = √(1-〖sin〗^2 α) = √(1- (√3/2)^2 ) = √(1- 3/4)  = √(1/4)  = 1/2 ∴	tanα = sinα/cosα = (√3/2  )/(1/2) = √3  Again,	cosβ = √3/2  ∴	sinβ = √(1-〖cos〗^2 β)  = √(1- (√3/2)^2 ) = √(1- 3/4)  = √(1/4)  = 1/2 ∴	tanβ = sinβ/cosβ = (1/2  )/(√3/2) = 1/√3 Now,  (tan⁡α  -〖 tan〗⁡β)/(1 +〖 tan〗⁡α  tan⁡β ) = (√3  - 1/√3)/(1 + √3   .  1/√3) = ((3 - 1)/√3)/(1 + 1) = (2/√3)/2 = 1/√3   proved.

 

Example 7: If tanA + cotA = 3, prove that tan2A + cot2A = 7.

Solution: Here,

          tanA + cotA = 3

          Squaring both sides, we get

          tan2A + 2tanA.cotA + cot2A = 9

or,     tan2A + 2×1 + cot2A = 9   [ tanA × cotA = 1]

or,     tan2A + 2 + cot2A = 9

or,     tan2A + cot2A = 9 – 2

or,     tan2A + cot2A = 7.  Proved.

 

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