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.
Hey! Looking for some great resources suitable for young ones? You've come to the right place. Check out our self-paced courses designed for students of grades I-XII.
Start with topics like Python, HTML, ML, and learn to make some games and apps all with the help of our expertly designed content! So students worry no more, because GeeksforGeeks School is now here!
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)