If \(\sin \theta = \frac{3}{5}\), what is the value of \(\cos \theta\)?
Answer: B
We use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\).
\((\frac{3}{5})^2 + \cos^2 \theta = 1\)
\(\frac{9}{25} + \cos^2 \theta = 1\)
\(\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}\)
\(\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}\) (Assuming θ is in the first quadrant).
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What is the value of \(\tan 45°\)?
Answer: B
\[ \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1 \]
Geometrically, a 45° right triangle is isosceles, so the opposite and adjacent sides are equal.
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The expression \(\sin^2 A + \cos^2 A\) is equal to:
Answer: B
This is the fundamental Pythagorean identity.
\[ \sin^{2}A + \cos^{2}A = 1 \]
It follows from Pythagoras applied to a right triangle of hypotenuse 1, and it holds for every angle \( A \).
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If a tree casts a shadow of 10 meters when the angle of elevation of the sun is 60°, what is the height of the tree?
Answer: B
Let 'h' be the height of the tree and 's' be the length of the shadow. The angle of elevation is θ.
We have \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{s}\).
\(\tan 60° = \frac{h}{10}\)
Since \(\tan 60° = \sqrt{3}\), we have \(\sqrt{3} = \frac{h}{10}\).
h = 10√3 meters.
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What is the value of \(\sec^2 \theta - \tan^2 \theta\)?
Answer: B
This is another key Pythagorean identity. It is derived from \(\sin^2 \theta + \cos^2 \theta = 1\) by dividing the entire equation by \(\cos^2 \theta\).
\(\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}\)
\(\tan^2 \theta + 1 = \sec^2 \theta\)
Rearranging gives \(\sec^2 \theta - \tan^2 \theta = 1\).
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If \(\tan \theta = 1\), what is the value of \(\sin \theta \cos \theta\)?
Answer: A
If \(\tan \theta = 1\), then \(\theta = 45°\).
We need to find the value of \(\sin 45° \cos 45°\).
\(\sin 45° = \frac{1}{\sqrt{2}}\) and \(\cos 45° = \frac{1}{\sqrt{2}}\).
So, \(\sin \theta \cos \theta = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2}\).
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From the top of a 100 m high lighthouse, the angle of depression of a boat is 30°. The distance of the boat from the foot of the lighthouse is:
Answer: B
The angle of depression from the top of the lighthouse to the boat is equal to the angle of elevation from the boat to the top of the lighthouse.
Let the distance of the boat from the foot of the lighthouse be 'd'.
\(\tan 30° = \frac{\text{height}}{\text{distance}} = \frac{100}{d}\)
\(\frac{1}{\sqrt{3}} = \frac{100}{d}\)
d = 100√3 meters.
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The value of \(\cos 0° + \sin 90° + \tan 0°\) is:
Answer: C
\[ \cos 0^{\circ} = 1, \qquad \sin 90^{\circ} = 1, \qquad \tan 0^{\circ} = 0 \]
\[ 1 + 1 + 0 = 2 \]
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If \(\sec \theta + \tan \theta = x\), then \(\sec \theta\) is equal to:
Answer: A
We know that \(\sec^2 \theta - \tan^2 \theta = 1\), which can be factored as \((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\).
Given \(\sec \theta + \tan \theta = x\), we have \(\sec \theta - \tan \theta = \frac{1}{x}\).
Now we have a system of two linear equations:
1) \(\sec \theta + \tan \theta = x\)
2) \(\sec \theta - \tan \theta = 1/x\)
Adding the two equations gives: \(2\sec \theta = x + \frac{1}{x} = \frac{x^2+1}{x}\).
Therefore, \(\sec \theta = \frac{x^2+1}{2x}\).
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What is the maximum value of \(3\sin \theta + 4\cos \theta\)?
Answer: C
For an expression of the form \(a\sin \theta + b\cos \theta\), the maximum value is \(\sqrt{a^2+b^2}\) and the minimum value is \(-\sqrt{a^2+b^2}\).
Here, a = 3 and b = 4.
Maximum value = \(\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\).
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