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Some of the trignometric Questions I could not perform and my attempts in general :

  • $\cos \alpha = \cos \beta\cdot\cos\phi = \cos\gamma\cdot\cos\theta$ and $\sin \alpha = 2\sin\dfrac\phi2\cdot\sin\dfrac\theta2$

    To prove. $\tan\dfrac\alpha2=\pm \tan\dfrac\beta2\cdot \tan\dfrac\gamma2$

    I tried to write $\cos\phi$ and $\cos\theta$ from first equation and put in second.

  • $\cos^2\theta = \dfrac{m^2-1}{3}$ and $\tan\alpha = \tan^3\dfrac\theta2$

    To prove. $\sin^{2/3}\alpha + \cos^{2/3}\alpha = \left(\dfrac4{m^2}\right)^{1/3}$

    Tried to convert second equation into first.

  • $\dfrac{\cos(\alpha-3\theta)}{\cos^3\theta} = \dfrac{\sin(\alpha-3\theta)}{\sin^3\theta} = m$

    To show. $\cos\alpha = \dfrac{2-m^2}m$

    Tried to expand the equation also for below question.

  • $\lambda \cos2\theta = \cos(\theta + \alpha) and \lambda \sin2\theta = 2\sin(\theta + \alpha)$

    To eliminate $\theta$

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For the first problem, clearly, we need to eliminate $\theta,\phi$

Now, $\cos\phi=\frac{\cos\alpha}{\cos\beta}$ and $\cos\theta=\frac{\cos\alpha}{\cos\gamma}$

$$\text{and }\sin^2\alpha=\left(2\sin\dfrac\phi2\cdot\sin\dfrac\theta2\right)^2=(2\sin^2\dfrac\phi2)(2\sin^2\dfrac\theta2)=(1-\cos\phi)(1-\cos\theta)$$

$$\implies 1-\cos^2\alpha=\left(1-\frac{\cos\alpha}{\cos\beta}\right)\left(1-\frac{\cos\alpha}{\cos\gamma}\right)$$

$$\implies \cos^2\alpha(1+\cos\beta\cos\gamma)=\cos\alpha(\cos\beta+\cos\gamma)$$

Assuming $\cos\alpha\ne0, \cos\alpha=\frac{\cos\beta+\cos\gamma}{1+\cos\beta\cos\gamma}$

Applying Componendo and dividendo, $$\frac{1-\cos\alpha}{1+\cos\alpha}=\frac{1+\cos\beta\cos\gamma-(\cos\beta+\cos\gamma)}{1+\cos\beta\cos\gamma+\cos\beta+\cos\gamma}=\frac{(1-\cos\beta)(1-\cos\gamma)}{(1+\cos\beta)(1+\cos\gamma)}$$

As $\cos2A=\frac{1-\tan^2A}{1+\tan^2A}, \tan^2A=\frac{1-\cos2A}{1+\cos2A}$

$$\implies \tan^2\frac\alpha2=\tan^2\frac\beta2\cdot\tan^2\frac\gamma2 $$


For the second question, $$\tan\alpha= \tan^3\frac\theta2$$ $$\implies \frac{\sin\alpha}{\sin^3\frac\theta2}=\frac{\cos \alpha}{\cos^3\frac\theta2}=\pm\sqrt{\frac{\sin^2\alpha+\cos^2\alpha}{\left(\sin^3\frac\theta2\right)^2+\left(\cos^3\frac\theta2\right)^2}}$$

$$\text{Now,} \left(\sin^3\frac\theta2\right)^2+\left(\cos^3\frac\theta2\right)^2=\left(\sin^2\frac\theta2\right)^3+\left(\cos^2\frac\theta2\right)^3$$

$$=\left(\frac{1-\cos\theta}2\right)^3+\left(\frac{1+\cos\theta}2\right)^3 \text{ as }\cos2A=1-2\sin^2=2\cos^2A-1$$

$$=\frac{2(1+3\cos^2\theta)}8=\frac{m^2}4 \text{ as }\cos^2\theta=\frac{m^2-1}3$$

$$\implies \frac{\sin\alpha}{\sin^3\frac\theta2}=\frac{\cos \alpha}{\cos^3\frac\theta2}=\pm \frac2m$$

$$\implies \sin^3\frac\theta2=\pm\frac{m\sin\alpha}2\implies \sin\frac\theta2=\left(\pm\frac{m\sin\alpha}2\right)^{\frac13}$$

Similarly, find $\cos \frac\theta2$ and use $\cos^2\frac\theta2+\sin^2\frac\theta2=1$ to eliminate $\theta$

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  • $\begingroup$ you can remove the other. $\endgroup$ – RE60K May 1 '15 at 11:53
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Quickly on the first identity: try to prove it by using $\cos(\frac{\alpha}{2})=\cos(\alpha-\frac{\alpha}{2})=\cos(\alpha)\cos(\frac{\alpha}{2})+ \sin(\alpha)\sin(\frac{\alpha}{2})$ and similarly for $\sin(\frac{\alpha}{2})$. Then you can use the given identities involving $\cos(\beta),\dots,\sin(\frac{\theta}{2})$ to arrive at the thesis.

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