Trigonometric Ratios of Complementary Angles

Complementary angles and their trigonometric ratios:

We know from geometry if the sum of two angles is 90°, then one angle is called the complement of the other.

Two angles A and B are complementary if A + B = 90°. So, B = 90° - A.

For example, as 30° + 60° = 90°, 60° is called the complement of 30° and conversely, 30° is called the complement of 60°. 

Thus 27° is the complement of 60°; 43.5° is the complement of 46.5° etc.

Thus in general, (90° - θ) and θ are complementary angles. Trigonometric ratios of (90° - θ) are convertible to trigonometric ratios of θ.

Trigonometric Ratios of 90° - θ in Terms of Trigonometric ratios of θ

Let us see how we can find the trigonometrical ratios of 90° - θ, if we know those of θ°. 

Let PQR be a right-angled triangle in which ∠Q is the right angle.

Complementary Angles and their Trigonometric Ratios

Let ∠PRQ = θ. Then, ∠QPR = 180° - (90° + θ) = 90° - θ.

1. sin (90° - θ) = cos θ

Here, sin (90° - θ) = QRPR and cos θ = QRPR

Therefore, sin (90° - θ) = cos θ.


2. cos (90° - θ) = sin θ

Here, cos (90° - θ) = PQPR and sin θ = PQPR

Therefore, cos (90° - θ) = sin θ.


3. tan (90° - θ) = cot θ

Here, tan (90° - θ) = QRPQ and cot θ = QRPQ

Therefore, tan (90° - θ) = cot θ.


4. csc (90° - θ) = sec θ

Here, csc (90° - θ) = PRQR and sec θ = PRQR

Therefore, csc (90° - θ) = sec θ


5. sec (90° - θ) = csc θ

Here, sec (90° - θ) = PRPQ and csc θ = PRPQ

Therefore, sec (90° - θ) = csc θ.


6. cot (90° - θ) = tan θ

Here, cot (90° - θ) = PQQR and tan θ = PQQR

Therefore, cot (90° - θ) = tan θ.


Thus, we have the following conversions of trigonometric ratios of (90° - θ) in terms of trigonometric ratios of θ.

sin (90° - θ) = cos θ

cos (90° - θ) = sin θ

tan (90° - θ) = cot θ

cot (90° - θ) = tan θ

sec (90° - θ) = csc θ

csc (90° - θ) = sec θ

For example, cos 37° can be expressed as sine of the complementary angle of 37° because

                                cos 37° = cos (90° - 53°) = sin 53°.


Note: The measure of an angle can be expressed in degrees (°) as well as in radians. The measure of an angle is π radians (where π is 3.14, approximately) if its measure in degrees is 180°. Thus, 180° = π radians. This is also written as 180° = π.

Therefore, 1° = π180

              30° = π6

              45° = π4

              60° = π3

              90° = π2, etc.


Therefore, we can write sin (90° - β) = sin (π2 – β) = cos β

                                   cos (90° - β) = cos (π2 – β) = sin β

                                   tan (90° - β) = tan (π2 – β) = cot β

                                   csc (90° - β) = csc (π2 – β) = sec β

                                   sec (90° - β) = sec (π2 – β) = csc β

                                   cot (90° - β) = cot (π2 – β) = tan β.

Trigonometric Ratios of Complementary Angles

The values of trigonometrical ratios of 30° and 60°, which are complementary angles are compared below. This will help us to have a clear understanding of the relations shown before. 

sin 30° = cos 60° = 12

cos 30° = sin 60° = 32

tan 30° = cot 60° = 33

csc 30° = sec 60° = 2

sec 30° = csc 60° = 233

cot 30° = tan 60° = 3


Similarly, from the complementary angles formulae we get

sin 45° = cos 45° = 22

tan 45° = cot 45° = 1

csc 45 = sec 45° = 2

tan 45° = cot 45° = 1

Again, 

sin 90° = cos 0° = 1

cos 90° = sin 0° = 0


Problems on Trigonometric Ratios of Complementary Angles

Problems on evaluation using trigonometric ratios of complementary angles

1. Evaluate without using trigonometric table: sin25°2cos65°

Solution:

sin25°2cos65°

= sin25°2cos(90°25°)

= sin25°2sin25°; [since, cos (90° - θ) = sin θ]

12.



2. Evaluate without using trigonometric table: tan 38° ∙ tan 52°

Solution:

tan 38° ∙ tan 52°

= tan 38° ∙ tan (90° - 38°)

= tan 38° ∙ cot 38°; [Since, tan (90° - θ) = cot θ]

= tan 38° ∙ 1tan38°

= 1.


3. Evaluate without using trigonometric table: sin67°cos23° - sec12°csc78°

Solution:

sin67°cos23° - sec12°csc78°

= sin67°cos(90°67°) - sec12°csc(90°12°)

= sin67°cos(90°67°) - sec12°csc(90°12°)

= sin67°sin67° - sec12°sec12°

[Since, cos (90° - θ) = sin θ and csc (90° - θ) = sec θ]

= 1 - 1

= 0.


4. If cos 39° = xx2+y2, what is the value of tan 51°?

Solution:

Given that cos 39° = xx2+y2

Therefore, sin2 39° = 1 -  x2x2+y2

                            = x2+y2x2x2+y2

                            = y2x2+y2

Therefore, sin 39° = yx2+y2, (negative value is not acceptable)

Now, tan 51° = tan (90° - 39°)

                    =  cot 39°

                    = cos39°sin39°

                    = cos 39° ÷ sin 39°

                    = xx2+y2 ÷ yx2+y2

                    = xy.



5. If cos 37° = x then find the value of tan 53°.

Solution:

tan 53°

= tan (90° - 37°)

= cot 37°; [Since, tan (90° - θ) = cot θ]

= cos37°sin37°

= xsin37° ................ (i)


Now, sin2 37° = 1 - cos2 37°; [since, 1 - cos2 θ = sin2 θ]

Therefore, sin 37° = 1cos237°

                           = 1x2

Therefore, from (i), tan 53° = x1x2.


6. If sec  ϕ = csc  β and 0° < (ϕ, β) < 90°, find the value of sin (ϕ + β).

Solution:

sec  ϕ = csc  β

1cosϕ = 1sinβ

⟹ cos ϕ = sin β

⟹ cos ϕ = cos (90° - β)

⟹ ϕ = 90° - β

⟹ ϕ + β = 90°

Therefore, sin (ϕ + β) = sin 90° = 1.


7. Find the value of sin2 15° + sin2 25° + sin2 33° + sin2 57° + sin2 65° + sin2 75°.

Solution:

sin2 (90° - 75°) + sin2 (90° - 65°) + sin2 (90° - 57°) + sin2 57° + sin2 65° + sin2 75°.

= cos2 75° + cos2 65° + cos2 57° + sin2 57° + sin2 65° + sin2 75°.

= (sin2 57° + cos2 75°) + (sin2 65° + cos2 65°) + (sin2 57° + cos2 57°)

= 1 + 1 + 1; [Since, sin2 θ + cos2 θ = 1]

= 3.


8. If tan 49° ∙ cot (90° - θ) = 1, find θ.

Solution:

tan 49° ∙ cot (90° - θ) = 1

⟹ tan 49° ∙ tan θ = 1; [Since, cot (90° - θ) = tan θ]

⟹ tan θ = 1tan49°

⟹ tan θ = cot 49°

⟹ tan θ = cot (90° - 41°)

⟹ tan θ = tan 41°

⟹ θ = 41°

Therefore, θ = tan 41°.


Problems on establishing equality using trigonometric ratios of complementary angles

9. Prove that sin 33° cos 77° = cos 57° sin 13°

Solution:

LHS = sin 33° cos 77°

       = sin (90° - 57°) cos (90° - 13°)

       = cos 57° sin 13°

       = RHS. (Proved).


10. Prove that tan 11° + cot 63° = tan 27° + cot 79°

Solution:

LHS = tan 11° + cot 63°

       = tan (90° - 79°) + cot (90° - 27°)

       = cot 79° + tan 27°

       = tan 27° + cot 79°

       = RHS. (Proved).


Problems on establishing identities and simplification using trigonometric ratios of complementary angles

11. If P and Q are two complementary angles, show that

(sin P + sin Q)2 = 1 + 2 sin P cos P

Solution:

Since P are Q are complementary angles,

Therefore, sin Q = sin (90° - P) = cos P

Therefore, (sin P + sin Q)2 = (sin P + cos P)2 

                                       = sin2 P + cos2 P + 2 sin P cos P

                                       = (sin2 P + cos2 P) + 2 sin P cos P

                                       = 1 + 2 sin P cos P


12. Simplify: sin(π2θ)cot(π2θ)sinθ

Solution: 

sin(π2θ)cot(π2θ)sinθ

cosθtanθsinθ, [Since sin (π2 - θ) = sin (90° - θ) = cos θ and cot (π2 - θ) = cot (90° - θ) = tan θ]

= cosθsinθcosθsinθ

sinθsinθ

= 1.


13. Prove that, sin2 7° + sin2 83°

Solution:

sin 83° = sin (90° - 7°) 

           = cos 7°; [since, sin (90° - θ) = cos θ]

LHS = sin2 7° + sin2 83°

      = sin2 7° + cos2 7°, [Since, sin 83° = cos 7°]

      = 1 = RHS (Proved).


14. In a ∆PQR, prove that sin P+Q2 = cos R2.

Solution:

We know that sum of the three angles of a triangle is 180°.

i,e., P + Q + R = 180°

⟹  P + Q = 180° - R

Now, 

LHS = sin P+Q2 

      = sin 180°R2 

      = sin (90° - R2)

      = cos R2 = RHS (Proved). 


15. Prove that tan 15° + tan 75° = sec215°sec215°1.

Solution:

LHS = tan 15° + tan (90° - 15°)

       = tan 15° + cot 15°

       = tan 15° + 1tan15°

       = tan215°+1tan15°

       = sec215°sec215°1 = RHS (Proved).


Learn more about Trigonometrical Ratios of Complementary Angles.





10th Grade Math

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