Trigonometry Questions and Answers

Q: If cosθ=−3/5 then find tanθ, where θ is an angle in the second quadrant:

\(\tan\theta = -\frac{4}{3}\).From \(\cos\theta = -\frac{3}{5}\), use \(\sin^{2}\theta + \cos^{2}\theta = 1\):\[ \begin{aligned} \sin^{2}\theta &= 1 - \frac{9}{25} = \frac{16}{25} \\ \sin\theta &= \frac{4}{5} \\ \tan\theta &= \frac{4/5}{-3/5} = -\frac{4}{3} \end{aligned} \]Sine is positive in the second quadrant, which is why \(\sin\theta\) is \(+\frac{4}{5}\), and tangent, sine over cosine, is negative there. That rules out \(\frac{4}{3}\). \(-\frac{4}{5}\) is a value of sine with the wrong sign, and \(\frac{3}{5}\) is cosine without its sign.

Q: What is the value of (sin15°+cos15°)2+(sin15°-cos15°)2 ?

The value is 2.Expand both squares; the cross terms cancel:\[ \begin{aligned} (s + c)^{2} + (s - c)^{2} &= 2s^{2} + 2c^{2} \\ &= 2(\sin^{2}15^\circ + \cos^{2}15^\circ) \\ &= 2 \times 1 = 2 \end{aligned} \]Here \(s = \sin 15^\circ\) and \(c = \cos 15^\circ\). The \(+2sc\) from the first square and the \(-2sc\) from the second cancel, so the angle does not matter at all. 1 forgets the factor 2: it is the value of \(\sin^{2} + \cos^{2}\) alone. 0 and \(-1\) are impossible, since a sum of two squares cannot be zero or negative here.

Important Instructions

101.

If cosθ=−3/5 then find tanθ, where θ is an angle in the second quadrant:

A.-4/3
B.-4/5
C.4/3
D.3/5

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: A

\(\tan\theta = -\frac{4}{3}\).

From \(\cos\theta = -\frac{3}{5}\), use \(\sin^{2}\theta + \cos^{2}\theta = 1\):

\[ \begin{aligned} \sin^{2}\theta &= 1 - \frac{9}{25} = \frac{16}{25} \\ \sin\theta &= \frac{4}{5} \\ \tan\theta &= \frac{4/5}{-3/5} = -\frac{4}{3} \end{aligned} \]

Sine is positive in the second quadrant, which is why \(\sin\theta\) is \(+\frac{4}{5}\), and tangent, sine over cosine, is negative there. That rules out \(\frac{4}{3}\). \(-\frac{4}{5}\) is a value of sine with the wrong sign, and \(\frac{3}{5}\) is cosine without its sign.

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102.

What is the value of (sin15°+cos15°)²+(sin15°-cos15°)² ?

A.1
B.2
C.0
D.-1

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: B

The value is 2.

Expand both squares; the cross terms cancel:

\[ \begin{aligned} (s + c)^{2} + (s - c)^{2} &= 2s^{2} + 2c^{2} \\ &= 2(\sin^{2}15^\circ + \cos^{2}15^\circ) \\ &= 2 \times 1 = 2 \end{aligned} \]

Here \(s = \sin 15^\circ\) and \(c = \cos 15^\circ\). The \(+2sc\) from the first square and the \(-2sc\) from the second cancel, so the angle does not matter at all. 1 forgets the factor 2: it is the value of \(\sin^{2} + \cos^{2}\) alone. 0 and \(-1\) are impossible, since a sum of two squares cannot be zero or negative here.

Discuss in Forum

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