If a sin θ – b cos θ = c then prove that a cos θ + b sin θ = ± √(a2 + b2 – c2).
Trigonometry is basically the study of the relationship between the angles and the sides of a triangle. It is one of the widely used topics of mathematics that is used in daily life. It involves operations on a right-angled triangle i.e. a triangle having one of the angles equal to 90°. There are some terms that we should know before going further. These terms are,
- Hypotenuse – It is the side opposite to the right angle in a right-angled triangle. It is the longest side of a right-angled triangle. In figure1 side, AC is the hypotenuse.
- Perpendicular – The perpendicular of a triangle, corresponding to a particularly acute angle θ is the side opposite to the angle θ. In figure1 side, AB is the perpendicular corresponding to angle θ.
- Base – It is the side adjacent to a particularly acute angle θ. In figure 1 side BC is the base corresponding to angle θ.
As earlier said, trigonometry depicts the relationship between the angles and sides of a right-angled triangle. This relationship is represented by standard ratios and is given as follows:
- Sine (sin): The sine of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the hypotenuse of the triangle.
sin θ = perpendicular/hypotenuse = p/h
- Cosine (cos): The cosine of an angle θ is the ratio of the length of the base, corresponding to the angle θ, to the length of the hypotenuse of the triangle.
cos θ = base/hypotenuse = b/h
- Tangent (tan): The tangent of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the base for the particular angle of the triangle.
tan θ = perpendicular/base = p/b
- Cotangent (cot): it is the reciprocal of a tangent.
cot θ = 1/tan θ = base/perpendicular = b/p
- Secant (sec): it is the reciprocal of cosine.
sec θ = 1/cos θ = hypotenuse/base = h/b
- Cosecant (cosec): it is the reciprocal of sine.
cosec θ = 1/sin θ = hypotenuse/perpendicular = h/p
Trigonometric functions of Complementary angles
One of the relations of trigonometry includes the concept of complementary angles. Complementary angles are a set of two angles, say x and y, such that on adding them they evaluate to 90°. Therefore, we can say x = 90° – y. There exist a special complementary relation between the trigonometric ratios as given below
Between sin and cos:
sin(90° – x) = cos x
cos(90° – x) = sin x
Between tan and cot:
tan(90° – x) = cot x
cot(90° – x) = tan x
Between sec and cosec:
sec(90° – x) = cosec x
cosec(90° – x) = sec x
These identities depend on the Pythagorean theorem. Applying Pythagoras theorem to the right-angled triangle, we get:
Opposite2 + Adjacent2 = Hypotenuse2
Dividing both sides by Hypotenuse2
Opposite2/Hypotenuse2 + Adjacent2/Hypotenuse2 = Hypotenuse2/Hypotenuse2
sin2x + cos2x = 1
1 + tan2x = sec2x
1 + cot2x = cosec2x
In the question given below, some complementary relationships between the trigonometric ratios will be used.
If a sin θ – b cos θ = c , prove that a cos θ + b sin θ = ± √(a2 + b2 – c2)
a sin θ – b cos θ = c
Squaring both sides:
⇒(a sin θ – b cos θ)2 = c2
⇒a2sin2θ + b2cos2θ – 2ab sinθ cosθ = c2
Using the property that sin2θ = 1-cos2θ and cos2θ = 1-sin2θ
⇒a2(1-cos2θ) + b2(1-sin2θ) – 2ab sinθ cosθ = c2
⇒a2 – a2cos2θ + b2 – b2sin2θ – 2ab sinθ cosθ = c2
Moving a2 and b2 to right hand side
⇒-a2cos2θ – b2sin2θ – 2ab sinθ cosθ = c2-a2-b2
Multiplying by -1 on both sides
⇒a2cos2θ + b2sin2θ + 2ab sinθ cosθ = a2+b2-c2
Now if we observe left hand side is square of (a cosθ + b sin θ)
⇒(a cosθ + b sin θ)2 = a2+b2-c2
Taking Square roots on both sides
a cosθ + b sin θ = ± √(a2 + b2 – c2)
Question 1: If θ be an acute angle and 7sin2θ + 3cos2θ = 4, then find the value of tanθ.
7sin2θ + 3cos2θ = 4
7sin2θ + 3(1-sin2θ) = 4
7sin2θ + 3 – 3sin2θ = 4
4sin2θ = 1
sin2θ = 1/4
sinθ = 1/2
So, θ = 30o
then tanθ = 1/√3
Question 2: If cosα = a cosβ and sinα = b sinβ, then find the value of sin2β in terms of a and b.
On squaring both sides
cos2α = a2 cos2β
=> 1 – sin2α = a2(1 – sin2β) ……..(1)
Again, sinα = b sinβ
On squaring both sides
sin2α = b2 sin2β
Put the value of sin2α in (1)
1 – b2 sin2β = a2 – a2sin2β)
a2 – 1 = a2sin2β – b2 sin2β
a2 – 1 = sin2β(a2 – b2)
sin2β = a2 – 1 / (a2 – b2)
Question 3: a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a2 + b2 + c2 = ab + bc + ca, then find the value of (sin2A + sin2B + sin2C).
Acc. to question
a2 + b2 + c2 – ab – bc – ca = 0
=>2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0
=>(a-b)2 + (b-c)2 + (c-a)2 = 0
All three sides are equal then it is equilateral triangle.
then ∠A = ∠B = ∠C = 60°
So, sin260 + sin260 + sin260
Question 4: If cot θ = 7/8, evaluate
(i) ((1 + sinθ) * (1 – sinθ))/(1 + cosθ) * (1 – cosθ)))
(i) Using (a + b) * (a – b) = a2 – b2 in numerator and denominator
(1 – sin2θ)/(1 – cos2θ)
Using sin2θ + cos2θ = 1
cos2θ/sin2θ = cot2θ
cot2θ = (7/8)2 = 49/64
(ii) cot2θ = (7/8)2 = 49/64
Question 5: If 3 cot A = 4, Check whether (1 – tan2A)/(1 + tan2A) = cos2A – sin2A
We know that, tanA = sinA / cosA ….(1)
Using (1) on L.H.S
= (1 – sin2A/cos2A)/(1 + sin2A/cos2A)
which on rearranging becomes
= (cos2A – sin2A)/(cos2A + sin2A)
Using the identity,
cos2A + sin2A = 1
= (cos2A – sin2A)
This is equal to RHS.
LHS = RHS (for every value of cot A)
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