Some trigonometric solutions based problems on trigonometric ratios are shown here with the stepbystep explanation.
1. If sin θ = 8/17, find other trigonometric ratios of <θ.
Solution:
Let us draw a ∆ OMP in which ∠M = 90°.
Then sin θ = MP/OP = 8/17.
Let MP = 8k and OP = 17k, where k is positive.
By Pythagoras’ theorem, we get
⇒ OM = 15k
Therefore, sin θ = MP/OP = 8k/17k = 8/17
cos θ = OM/OP = 15k/17k = 15/17
tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15
csc θ = 1/sin θ = 17/8
sec θ = 1/cos θ = 17/15 and
cot θ = 1/tan θ = 15/8.
2. If Cos A = 9/41, find other trigonometric ratios of ∠A.
Solution:
Let us draw a ∆ ABC in which ∠B = 90°.
Then cos θ = AB/AC = 9/41.
Let AB = 9k and AC = 41k, where k is positive.
By Pythagoras’ theorem, we get
AC^{2} = AB^{2} + BC^{2}⇒ BC = 40k
Therefore, sin A = BC/AC = 40k/41k = 40/41
cos A = AB/AC = = 9k/41k = 9/41
tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9
csc A = 1/sin A = 41/40
sec A = 1/cos A = 41/9 and
cot A = 1/tan A = 9/40.
3. Show that the value of sin θ and cos θ cannot be more than 1.
Solution:
We know, in a right angle triangle the hypotenuse is the longest side.
sin θ = perpendicular/hypotenuse = MP/OP < 1 since perpendicular cannot be greater than hypotenuse; sin θ cannot be more than 1.
Similarly, cos θ = base/hypotenuse = OM/OP < 1 since base cannot be greater than hypotenuse; cos θ cannot be more than 1.
4. Is that possible when A and B be acute angles, sin A = 0.3 and cos B = 0.7?
Solution:
Since A and B are acute angles, 0 ≤ sin A ≤ 1 and 0 ≤ cos B ≤ 1, that means the value of sin A and cos B lies between 0 to 1. So, it is possible that sin A = 0.3 and cos B = 0.7
5. If 0° ≤ A ≤ 90° can sin A = 0.4 and cos A = 0.5 be possible?
Solution:
We know that sin^{2}A + cos^{2}A = 1Let us draw a ∆ ABC in which ∠B = 90° and ∠BAC = θ.
Then sin θ = BC/AC = 1/2.
Let BC = k and AC = 2k, where k is positive.
By Pythagoras’ theorem, we get
AC^{2} = AB^{2} + BC^{2}= 3√3/2  4 × 3√3/8
= 3√3/2  3√3/2
= 0
Hence, (3cos θ  4 cos<sup>3</sup> θ) = 0.
7. Show that sin α + cos α > 1 when 0° ≤ α ≤ 90°
Solution:
From the right triangle MOP,
Sin α = perpendicular/ hypotenuse
Cos α = base/ hypotenuse
Now, Sin α + Cos α
= perpendicular/ hypotenuse + base/ hypotenuse
= (perpendicular + base)/hypotenuse, which is > 1, Since we know that the sum of two sides of a triangle is always greater than the third side.
8. If cos θ = 3/5, find the value of (5csc θ  4 tan θ)/(sec θ + cot θ)
Solution:
Let us draw a ∆ ABC in which ∠B = 90°.
Let ∠A = θ°
Then cos θ = AB/AC = 3/5.
Let AB = 3k and AC = 5k, where k is positive.
By Pythagoras’ theorem, we get
AC^{2} = AB^{2} + BC^{2}⇒ BC = 4k
Therefore, sec θ = 1/cos θ = 5/3
tan θ = BC/AB =4k/3k = 4/3
cot θ = 1/tan θ = 3/4 and
csc θ = AC/BC = 5k/4k = 5/4
Now (5csc θ 4 tan θ)/(sec θ + cot θ)
= (5 × 5/4  4 × 4/3)/(5/3 + 3/4)
= (25/4 16/3)/(5/3 +3/4)
= 11/12 × 12/29
= 11/29
9. Express 1 + 2 sin A cos A as a perfect square.
Solution:
1 + 2 sin A cos A
= sin^{2} A + cos^{2} A + 2sin A cos A, [Since we know that sin^{2} θ + cos^{2} θ = 1]10. If sin A + cos A = 7/5 and sin A cos A =12/25, find the values of sin A and cos A.
Solution:
sin A + cos A = 7/5
⇒ cos A = 7/5  sin θ
Now from sin θ/cos θ = 12/25
We get, sin θ(7/5  sin θ) = 12/25
or, 7 sin θ – 5 sin^{2} θ = 12/5or, 5 sin θ(5 sin θ  4)  3(5 sin θ  4) = 0
or, (5 sin θ  3) (5 sin θ  4) = 0
⇒ (5 sin θ  3) = 0 or, (5 sin θ  4) = 0
⇒ sin θ = 3/5 or, sin θ = 4/5
When sin θ = 3/5, cos θ = 12/25 × 5/3 = 4/5
Again, when sin θ = 4/5, cos θ = 12/25 × 5/4 = 3/5
Therefore, sin θ =3/5, cos θ = 4/5
or, sin θ =4/5, cos θ = 3/5.
11. If 3 tan θ = 4, evaluate (3sin θ + 2 cos θ)/(3sin θ  2cos θ).
Solution: Given,
3 tan θ = 4
⇒ tan θ = 4/3
Now,
(3sin θ + 2 cos θ)/(3sin θ  2cos θ)
= (3 tan θ + 2)/(3 tan θ  2), [dividing both numerator and denominator by cos θ]
= (3 × 4/3 + 2)/(3 × 4/3 2), putting the value of tan θ = 4/3
= 6/2
= 3.
12. If (sec θ + tan θ)/(sec θ  tan θ) = 209/79, find the value of θ.
Solution: (sec θ + tan θ)/(sec θ  tan θ) = 209/79
⇒ [(sec θ + tan θ)  (sec θ  tan θ)]/[(sec θ + tan θ) + (sec θ  tan θ)] = [209 – 79]/[209 + 79], (Applying componendo and dividendo)
⇒ 2 tan θ/2 sec θ =130/288
⇒ sin θ/cos θ × cos θ = 65/144
⇒ sin θ = 65/144.
13. If 5 cot θ = 3, find the value of (5 sin θ  3 cos θ)/(4 sin θ + 3 cos θ).
Solution:
Given 5 cot θ = 3
⇒ cot θ = 3/5
Now (5 sin θ  3 cos θ)/(4 sin θ + 3 cos θ)
= (5  3 cot θ)/(4 sin θ + 3 cot θ), [dividing both numerator and denominator by sin θ]
= (5  3 × 3/5)/(4 + 3 × 3/5)
= (5  9/5)/(4 + 9/5)
= (16/5 × 5/29)
= 16/29.
⇒ sin θ(sin θ  2)  1(sin θ  2) = 0
⇒ (sin θ  2)(sin θ  1) = 0
⇒ (sin θ  2) = 0 or, (sin θ  1) = 0
⇒ sin θ = 2 or, sin θ = 1
So, value of sin θ can’t be greater than 1,
Therefore sin θ = 1
⇒ θ = 90°
`Relations Between the Trigonometric Ratios
Problems on Trigonometric Ratios
Reciprocal Relations of Trigonometric Ratios
Problems on Trigonometric Identities
Elimination of Trigonometric Ratios
Eliminate Theta between the equations
Verify Trigonometric Identities
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