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If a sin θ – b cos θ = c then prove that a cos θ + b sin θ = ± √(a2 + b2 – c2).

  • Last Updated : 26 Sep, 2021

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,

  1. 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.
  2. 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.
  3. Base – It is the side adjacent to a particularly acute angle θ. In figure 1 side BC is the base corresponding to angle θ.

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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



Trigonometric Indentities

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)

Solution:

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)

Hence proved!

Similar Problems

Question 1: If θ be an acute angle and 7sin2θ + 3cos2θ = 4, then find the value of tanθ.

Solution: 

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.

Solution: 

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).



Solution: 

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

=> a=b=c

All three sides are equal then it is equilateral triangle.

then ∠A = ∠B = ∠C = 60°

So, sin260 + sin260 + sin260

= 3(√3/2)2

= 9/4

Question 4: If cot θ = 7/8, evaluate

(i) ((1 + sinθ) * (1 – sinθ))/(1 + cosθ) * (1 – cosθ)))

(ii) cot2θ

Solution:

(i) Using (a + b) * (a – b) = a2 – b2 in numerator and denominator

We get

(1 – sin2θ)/(1 – cos2θ)

Using sin2θ + cos2θ = 1

We get

cos2θ/sin2θ = cot2θ

Now

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

Solution:

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

LHS becomes

= (cos2A – sin2A)

This is equal to RHS.

LHS = RHS (for every value of cot A)

Hence, Proved.




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