Introduction to Trigonometry | Exercise 8.1

Question 9

In triangle ABC, right-angled at B, if tan⁡A=13\tan A = \dfrac{1}{\sqrt{3}}, find the value of:

(i) sin⁡Acos⁡C+cos⁡Asin⁡C\sin A \cos C + \cos A \sin C

(ii) cos⁡Acos⁡C−sin⁡Asin⁡C\cos A \cos C - \sin A \sin C

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Solution
Understand the Question
  • Given tan⁡A=13=OppositeAdjacent\tan A = \dfrac{1}{\sqrt{3}} = \dfrac{\text{Opposite}}{\text{Adjacent}}, so for ∠A\angle A, let opposite side BC=k\text{BC} = k and adjacent side AB=3k\text{AB} = \sqrt{3}k, where k>0k > 0.
  • Use the Pythagoras theorem to find the hypotenuse AC\text{AC}.
  • Note that the reference angle changes between AA and CC:
    • For ∠A\angle A: Opposite side is BC\text{BC}, Adjacent side is AB\text{AB}.
    • For ∠C\angle C: Opposite side is AB\text{AB}, Adjacent side is BC\text{BC}.
  • Compute sin⁡A\sin A, cos⁡A\cos A, sin⁡C\sin C, and cos⁡C\cos C, then substitute them into each given expression.

(i) Find the value of sin⁡Acos⁡C+cos⁡Asin⁡C\sin A \cos C + \cos A \sin C

Step 1 · Find Sides of the Triangle and Trigonometric Ratios

Consider right triangle ABC\text{ABC} with ∠B=90∘\angle \text{B} = 90^\circ.Diagram 1

Given tan⁡A=Opposite sideAdjacent side=BCAB=13\tan A = \dfrac{\text{Opposite side}}{\text{Adjacent side}} = \dfrac{\text{BC}}{\text{AB}} = \dfrac{1}{\sqrt{3}}

Let BC=k\text{BC} = k and AB=3k\text{AB} = \sqrt{3}k for some positive number kk.

By Pythagoras theorem in ΔABC\Delta \text{ABC}

AC2=AB2+BC2=(3k)2+(k)2=3k2+k2=4k2AC=4k2=2k\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ &= (\sqrt{3}k)^2 + (k)^2 \\ &= 3k^2 + k^2 \\ &= 4k^2 \\ \text{AC} &= \sqrt{4k^2} = 2k \end{aligned}

Now, calculate the required trigonometric ratios for ∠A\angle A and ∠C\angle C

sin⁡A=BCAC=k2k=12cos⁡A=ABAC=3k2k=32\begin{aligned} \sin A &= \dfrac{\text{BC}}{\text{AC}} = \dfrac{k}{2k} = \dfrac{1}{2} \\[0.6em] \cos A &= \dfrac{\text{AB}}{\text{AC}} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2} \end{aligned}

For ∠C\angle C, AB\text{AB} is opposite and BC\text{BC} is adjacent

sin⁡C=ABAC=3k2k=32cos⁡C=BCAC=k2k=12\begin{aligned} \sin C &= \dfrac{\text{AB}}{\text{AC}} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2} \\[0.6em] \cos C &= \dfrac{\text{BC}}{\text{AC}} = \dfrac{k}{2k} = \dfrac{1}{2} \end{aligned}

Step 2 · Evaluate the Expression

Substitute the values of sin⁡A\sin A, cos⁡C\cos C, cos⁡A\cos A, and sin⁡C\sin C

sin⁡Acos⁡C+cos⁡Asin⁡C=(12)(12)+(32)(32)=14+34=1+34=44=1\begin{aligned} \sin A \cos C + \cos A \sin C &= \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \\[0.6em] &= \frac{1}{4} + \frac{3}{4} \\[0.6em] &= \frac{1+3}{4} \\[0.6em] &= \frac{4}{4} \\[0.6em] &= 1 \end{aligned}
Answer

(i) 11

(ii) Find the value of cos⁡Acos⁡C−sin⁡Asin⁡C\cos A \cos C - \sin A \sin C

Step 1 · Evaluate the Expression

Substitute the values of cos⁡A\cos A, cos⁡C\cos C, sin⁡A\sin A, and sin⁡C\sin C

cos⁡Acos⁡C−sin⁡Asin⁡C=(32)(12)−(12)(32)=34−34=0\begin{aligned} \cos A \cos C - \sin A \sin C &= \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right) - \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \\[0.6em] &= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\[0.6em] &= 0 \end{aligned}
Answer

(ii) 00

Common Mistakes
  • Reference Angle Confusion: The opposite and adjacent sides swap depending on whether you are considering ∠A\angle A or ∠C\angle C. For ∠A\angle A, side BC\text{BC} is opposite, but for ∠C\angle C, side AB\text{AB} is opposite.
  • Trigonometric Identities: Note that in right ΔABC\Delta \text{ABC} (with ∠B=90∘\angle B = 90^\circ), A+C=90∘A + C = 90^\circ. Hence sin⁡(A+C)=sin⁡90∘=1\sin(A+C) = \sin 90^\circ = 1 and cos⁡(A+C)=cos⁡90∘=0\cos(A+C) = \cos 90^\circ = 0.

More questions in Exercise 8.1

Q1

In ΔABC\Delta \text{ABC}, right-angled at B, AB=24 cm\text{AB} = 24\text{ cm}, BC=7 cm\text{BC} = 7\text{ cm}. Determine :

(i) sin⁡A\sin A, cos⁡A\cos A

(ii) sin⁡C\sin C, cos⁡C\cos C

Q2

In Fig. 8.13, find tan⁡P−cot⁡R\tan P - \cot R.

Q3

If sin⁡A=34\sin A = \dfrac{3}{4}, calculate cos⁡A\cos A and tan⁡A\tan A.

Q4

Given 15cot⁡A=815 \cot A = 8, find sin⁡A\sin A and sec⁡A\sec A.

Q5

Given sec⁡θ=1312\sec \theta = \dfrac{13}{12}, calculate all other trigonometric ratios.

Q6

If ∠A\angle A and ∠B\angle B are acute angles such that cos⁡A=cos⁡B\cos A = \cos B, then show that ∠A=∠B\angle A = \angle B.

Q7

If cot⁡θ=78\cot \theta = \dfrac{7}{8}, evaluate :

(i) (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}

(ii) cot⁡2θ\cot^2 \theta

Q8

If 3cot⁡A=43 \cot A = 4, check whether 1−tan⁡2A1+tan⁡2A=cos⁡2A−sin⁡2A\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A or not.

Q9

In triangle ABC, right-angled at B, if tan⁡A=13\tan A = \dfrac{1}{\sqrt{3}}, find the value of:

(i) sin⁡Acos⁡C+cos⁡Asin⁡C\sin A \cos C + \cos A \sin C

(ii) cos⁡Acos⁡C−sin⁡Asin⁡C\cos A \cos C - \sin A \sin C

Q10

In ΔPQR\Delta \text{PQR}, right-angled at Q, PR+QR=25 cm\text{PR} + \text{QR} = 25\text{ cm} and PQ=5 cm\text{PQ} = 5\text{ cm}. Determine the values of sin⁡P\sin P, cos⁡P\cos P and tan⁡P\tan P.

Q11

State whether the following are true or false. Justify your answer.

(i) The value of tan⁡A\tan A is always less than 1.

(ii) sec⁡A=125\sec A = \dfrac{12}{5} for some value of angle A.

(iii) cos⁡A\cos A is the abbreviation used for the cosecant of angle A.

(iv) cot⁡A\cot A is the product of cot⁡\cot and AA.

(v) sin⁡θ=43\sin \theta = \dfrac{4}{3} for some angle θ\theta.

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