Trigonometric Functions

 $x,y\in \left(0,\pi \right)$

cos x + cos y – cos (x + y) = $\frac{3}{2}$

$2cos\frac{x+y}{2}.cos\frac{x-y}{2}-\left(2co{s}^2\frac{x+y}{2}-1\right)=\frac{3}{2}$

$\Rightarrow 2co{s}^2\frac{x+y}{2}-2cos\frac{x-y}{2}.cos\frac{x+y}{2}+\frac{1}{2}=0$

As,

$x,y\in \left(0,\pi \right)$

$\frac{-\pi }{2}<\frac{x-y}{2}<\frac{\pi }{2}$

$sin\frac{x-y}{2}=0\Rightarrow \frac{x-y}{2}=0\Rightarrow x=y$

then equation : cos x + cos y – cos (x + y) = $\frac{3}{2}$

$cosx=\frac{2\pm \sqrt[]{4-4}}{4}=\frac{1}{2}$

$x=y=\frac{\pi }{3}\Rightarrow sinx+cosy=\frac{\sqrt[]{3}}{2}+\frac{1}{2}=\frac{\sqrt[]{3}+1}{2}$

$si{n}^{7}x=co{s}^{7}x=1,x\in \left[0,4\pi \right]$            

will satisfy for sin x = 1, cos x = 0

$\Rightarrow x=\frac{\pi }{2}\&\frac{5\pi }{2}.$              

or, cos x = 1, sin x = 0

x = 0, 2π, 4π              total 5 solutions

The parabola : ${\left(x-\frac{1}{2}\right)}^2=y-\frac{3}{4}$  

-> y = x² – x + 1 . (i)

$P\left(-\frac{1}{2},\frac{7}{4}\right)$                

$N_P:y=\frac{x}{2}+2.........\left(ii\right)$                

(i) & (ii) -> Q (2, 3)

$P{Q}^2=\frac{125}{16}$                

q = 18° Þ 2q + 3q = 90° Þ sin 3q = 1 – sin 2q Þ cos q Þ cosq (4 sin² q + 2sinq - 1) = 0

$?cos\theta \ne 0\therefore 4si{n}^2\theta +2sin\theta -1=0\Rightarrow cose{c}^{2}18°-2cosec18°-4=0$

? x² – 2x – 4 = 0

${\left(si{n}^{-1}x\right)}^2-{\left(co{s}^{-1}x\right)}^2=a,0<x<1$           

$\left(si{n}^{-1}x+co{s}^{-1}x\right)\left(si{n}^{-1}x-co{s}^{-1}x\right)=a$

$2co{s}^{-1}x=\frac{\pi }{2}-\frac{2a}{\pi }..........\left(i\right)$    

Let $co{s}^{-1}x=\theta thenx=cos\theta$

So, $2x^2-1=2co{s}^2\theta -1=cos2\theta ..........\left(ii\right)$

Now, $2\theta =2co{s}^{-1}x=\frac{\pi }{2}-\frac{2a}{\pi }from\left(i\right)$

So, cos 2θ = cos $\left(\frac{\pi }{2}-\frac{2a}{\pi }\right)=sin\frac{2a}{\pi }$

sin^4_θ + cos⁴θ - sinθ cosθ = 0

${\left(si{n}^2\theta +co{s}^2\theta \right)}^2-2si{n}^2\theta co{s}^2\theta -sin\theta cos\theta =0$

$\frac{{\left(2sin\theta cos\theta \right)}^2}{2}-\frac{2sin\theta cos\theta }{2}+1=0$               

$si{n}^{2}2\theta +sin2\theta -2=0$              

sin 2θ = 1 .(i)

$as\theta \in \left[0,4\pi \right]so2\theta \in \left[0,8\pi \right]$              

$\left(i\right)2\theta =\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2},\frac{18\pi }{2}$

Hence $\frac{8s}{\pi }=56$

In $\Delta BCD,tan?=\frac{x}{a+b}........\left(i\right)$  

In $\Delta APC,tan\left(\theta +?\right)=\frac{x}{b}..........\left(ii\right)$  

Now tan θ = tan $\left(\theta +?-?\right)$  

$=\frac{tan\left(\theta +?\right)-tan?}{1+tan\left(\theta +?\right)tan?}$  

given , $tan\theta =\frac{1}{2}so\frac{ax}{b\left(a+b\right)+x^2}=\frac{1}{2}$  

-> x² – 2ax + b (a + b) = 0

$\frac{sinA}{sinB}=\frac{sin\left(A-C\right)}{sin\left(C-B\right)}$

sin A sin C cos B – sin A cos C sin B = sin B sin A cos C – sin B cos A sin C

2 sin A sin B cos C = sin A sin C cos B + sin B sin C cos A

By sine rule $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k$

$2.ak.bk.\frac{\left(a^2+b^2-c^2\right)}{2ab}=ak.ck.\frac{\left(c^2+a^2-b^2\right)}{2ac}+bk.ck.\frac{\left(b^2+c^2-a^2\right)}{2bc}$

$\Rightarrow b^2,c^2,a^{2}areinA.P.$

  $ta{n}^{-1}\frac{1}{2r^2}=ta{n}^{-1}\frac{2}{1+\left(4r^2-1\right)}$             

  $=ta{n}^{-1}\frac{\left(2r+1\right)-\left(2r-1\right)}{1+\left(2r+1\right)\left(2r-1\right)}$

${\displaystyle \sum _{r=1}^{50}\left[ta{n}^{-1}\left(2r+1\right)-ta{n}^{-1}\left(2r-1\right)\right]}$

$=ta{n}^{-1}101-ta{n}^{-1}1=ta{n}^{-1}\frac{100}{102}$

$\Rightarrow tanP=\frac{100}{102}=\frac{50}{51}$             

$l=\underset{n\to x}{lim}\frac{1}{2^n}\left(\frac{1}{\sqrt[]{1-\frac{1}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{2}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{3}{2^n}}}+.....+\frac{1}{\sqrt[]{1-\frac{2^n-1}{2^n}}}\right)$

Let 2n = t and if n $\infty$ then t $\infty$

$l=\underset{n\to x}{lim}\frac{1}{t}\left({\displaystyle \sum _{r=1}^{t=1}\frac{1}{\sqrt[]{1+\frac{r}{t}}}}\right)$

$={\left[2x^{\frac{1}{2}}\right]}_0^1=2$