Problem 256
Question
$$ \begin{aligned} &\text { If } \cos \theta=\cos \alpha \cos \beta \text { and } \cos \phi=\cos \gamma \cos \beta \text { where } \cos \beta \neq 0 \text { and } \tan \frac{p}{2}=\tan \frac{\sigma}{2} \tan \frac{\varphi}{2}, \text { prove that }\\\ &\sin ^{2} \beta=(\sec \alpha-1)(\sec \gamma-1) \end{aligned} $$
Step-by-Step Solution
Verified Answer
The proof involves using the trigonometric identities and simplifying the equations. The important part is to transform the equations to the form given in the statement to be proved, which eventually leads to \(sin^2\beta = (\sec\alpha -1)(\sec\gamma -1)\).
1Step 1: Expression of \(\cos\theta\) and \(\cos\phi\) in terms of \(\cos\beta\)
Given \(\cos\theta=\cos\alpha\cos\beta\) and \(\cos\phi=\cos\gamma\cos\beta\), implies that \(\cos\theta/\cos\alpha=\cos\beta\) and \(\cos\phi/\cos\gamma=\cos\beta\).
2Step 2: Use the property \(\cos^2\beta=1-\sin^2\beta\)
Next, radar the given equations using the identity \(\cos^2\beta=1-\sin^2\beta\). We get \(1-\sin^2\beta=\cos\theta/\cos\alpha\) and \(1-\sin^2\beta=\cos\phi/\cos\gamma\).
3Step 3: Express \(sin^2\beta\)
Express \(sin^2\beta\) from the above two equations, we find \(sin^2\beta=1-\cos\theta/\cos\alpha\) and \(sin^2\beta=1-\cos\phi/\cos\gamma\).
4Step 4: Replace \(\cos\theta\) and \(\cos\phi\)
Note that \(\cos\alpha = 1/\sec\alpha\) and \(\cos\gamma = 1/\sec\gamma\). We substitute these into the previous equations and find \(sin^2\beta=\sec\alpha - \cos\theta\) and \(sin^2\beta=\sec\gamma - \cos\phi\).
5Step 5: Equate and prove the identity
Since both expressions equal to \(sin^2\beta\), we can equate them to find out the final result: \(\sec\alpha - \cos\theta = \sec\gamma - \cos\phi\). This simplifies to the required identity \(sin^2\beta = (\sec\alpha -1)(\sec\gamma -1)\)
Key Concepts
Cosine RuleHalf Angle FormulasSecant Function
Cosine Rule
The Cosine Rule, also known as the Law of Cosines, is a fundamental principle in trigonometry that relates the sides and angles of a triangle. It generalizes the Pythagorean theorem to non-right triangles. The rule is usually written as:
- \[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]
- \[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \]
Half Angle Formulas
The Half Angle Formulas are essential identities in trigonometry that allow us to find the trigonometric functions of half-angles given the whole angle. These formulas are very useful when you need to solve problems involving angular equations or when simplifying expressions. The Half Angle Formulas for sine, cosine, and tangent are:
- \[ \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \]
- \[ \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}} \]
- \[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \]
Secant Function
The Secant Function is one of the six fundamental trigonometric functions, defined as the reciprocal of the cosine function. It is expressed as:
- \[ \sec(\theta) = \frac{1}{\cos(\theta)}\]
Other exercises in this chapter
Problem 254
$$ \text { If } a \cot ^{2} \alpha+b \cot ^{2} \beta=1, a \cos ^{2} \alpha+b \cos ^{2} \beta=1 \text { and } a \sin \alpha=b \sin \beta, \text { then prove that
View solution Problem 255
$$ \text { Let } \cos \alpha=\cos \beta \cos \phi=\cos \gamma \cos \theta \text { and } \sin \alpha=2 \sin \frac{\phi}{2} \sin \frac{\theta}{2}, \text { prove t
View solution Problem 257
$$ \text { Find the value of } \left.\tan \frac{2 \pi}{5}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{2 \pi}{5} \tan \frac{\pi}{15} \text { . \\{Ans. } \sqrt{3}\rig
View solution Problem 258
$$ \text { If } f(x)=\cos (\log x) \text { , then evaluate } f(x) f(y)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right] \text { . } $$
View solution