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Trigonometry Help on Trigonometric sums. (Assorted type)

surajkrishna

New member
Jul 14, 2012
4
I have some questions and doubts in trigonometry. I hope somebody can solve these questions.

Q1) If for real values of x, cos[tex]\theta = x +\frac{1}{x}[/tex], then

a) [tex]\theta[/tex] is acute angle b) [tex]\theta[/tex] is right angle c) [tex]\theta[/tex] is an obtuse angle d) no value of [tex]\theta[/tex] is possible

I will post the following questions soon.
 

surajkrishna

New member
Jul 14, 2012
4
Q2) If sec[tex]\theta[/tex] + tan[tex]\theta[/tex] = x, then sin[tex]\theta[/tex] is equal to

a) x2+1/2x b) x2+1/x2-1 c) x2-1/x2+1 d) 2x/x2​-1
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have some questions and doubts in trigonometry. I hope somebody can solve these questions.

Q1) If for real values of x, cos[tex]\theta = x +\frac{1}{x}[/tex], then

a) [tex]\theta[/tex] is acute angle b) [tex]\theta[/tex] is right angle c) [tex]\theta[/tex] is an obtuse angle d) no value of [tex]\theta[/tex] is possible

I will post the following questions soon.
The function $\displaystyle f(x)=x+\frac{1}{x}$...

a) for x>0 f(x) is greater than 0 and it has a minimum for x=1, where is f(x)=2...

b) for x<0 f(x) is less than 0 and it has a maximum for x=-1, where is f(x)=-2...

Consequence of a) and b) is that it doesn't exist any real $\theta$ for which is $\displaystyle \cos \theta= x+\frac{1}{x}$...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Q2) If sec[tex]\theta[/tex] + tan[tex]\theta[/tex] = x, then sin[tex]\theta[/tex] is equal to

a) x2+1/2x b) x2+1/x2-1 c) x2-1/x2+1 d) 2x/x2​-1
$\displaystyle \frac{1}{\cos \theta}+ \frac{\sin \theta}{\cos \theta}=x \implies \frac{1+ \sin \theta}{\sqrt {1 - \sin^{2} \theta}}+ x \implies \sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=x \implies \sin \theta = \frac{x^{2}-1}{x^{2}+1}$

Kind regards

$\chi$ $\sigma$
 

surajkrishna

New member
Jul 14, 2012
4
$\displaystyle \frac{1}{\cos \theta}+ \frac{\sin \theta}{\cos \theta}=x \implies \frac{1+ \sin \theta}{\sqrt {1 - \sin^{2} \theta}}+ x \implies \sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=x \implies \sin \theta = \frac{x^{2}-1}{x^{2}+1}$

Kind regards

$\chi$ $\sigma$
$\displaystyle \frac{1+ \sin \theta}{\sqrt {1 - \sin^{2} \theta}}= x $

After this how do you get this:

$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=x \implies \sin \theta = \frac{x^{2}-1}{x^{2}+1}$
 

chisigma

Well-known member
Feb 13, 2012
1,704
$\displaystyle \frac{1+ \sin \theta}{\sqrt {1 - \sin^{2} \theta}}= x $

After this how do you get this:

$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=x \implies \sin \theta = \frac{x^{2}-1}{x^{2}+1}$

$\displaystyle \frac{1+ \sin \theta}{\sqrt {1 - \sin^{2} \theta}}= x \implies \sqrt{\frac{(1+\sin \theta)^{2}}{(1+\sin \theta)\ (1-\sin \theta)}} = x \implies \sqrt{\frac{1+ \sin \theta}{1-\sin \theta}}=x \implies 1 + \sin \theta= x^{2}\ (1-\sin \theta) \implies \sin \theta= \frac{x^{2}-1}{x^{2}+1} $

Kind regards

$\chi$ $\sigma$
 

surajkrishna

New member
Jul 14, 2012
4
Thanks for the help.