Trigonometric Functions

λ=- (sin? θ+cos? θ) = - (sin²θ+cos²θ)²-2sin²θcos²θ)
λ = - (1-½sin²2θ) = ½sin²2θ-1
sin²2θ ∈
λ ∈ [-1, -1/2]

Average speed = (f (t? )-f (t? )/ (t? -t? ) = a (t? +t? )+b.
Instantaneous speed = f' (t)=2at+b.
2at+b=a (t? +t? ) ⇒ t= (t? +t? )/2.

sinx+sin4x + sin2x+sin3x = 0
2sin (5x/2)cos (3x/2) + 2sin (5x/2)cos (x/2) = 0
2sin (5x/2) [cos (3x/2)+cos (x/2)] = 0
4sin (5x/2)cosxcos (x/2)=0.
sin (5x/2)=0 ⇒ 5x/2=kπ ⇒ x=2kπ/5. x=0, 2π/5, 4π/5, 6π/5, 8π/5, 2π.
cosx=0 ⇒ x=π/2, 3π/2.
cos (x/2)=0 ⇒ x=π.
Sum = 9π.

Let C be center, O be observer. Let R be radius of balloon. sin30° = R/OC = 16/OC ⇒ OC=32.
Let H be height of center. sin75°=H/OC ⇒ H=32sin75°.
Topmost point height = H+R = 32sin75°+16 = 8 (√6+√2)+16 = 8 (√6+√2+2).

L = sin (3π/16)sin (-π/16)
= (1/2) (cos (π/4) - cos (π/8)
= (1/2) (1/√2 - cos (π/8)
M = cos (3π/16)cos (-π/16)
= (1/2) (cos (π/4) + cos (π/8)
= (1/2) (1/√2 + cos (π/8)

cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2

sinθ + cosθ = 1/2
16 (sin (2θ) + cos (4θ) + sin (6θ)
= 16 [2sin (4θ)cos (2θ) + cos (4θ)]
= 16 [4sin (2θ)cos² (2θ) + 2cos² (2θ) - 1] . (i)
Now, sinθ + cosθ = 1/2, squaring on both sides, we get
1 + sin (2θ) = 1/4
sin (2θ) = -3/4
cos² (2θ) = 1 - sin² (2θ) = 1 - 9/16 = 7/16
From equation (i)
16 [4 (-3/4) (7/16) + 2 (7/16) - 1]
16 [-21/16 + 14/16 - 16/16] = 16 [-23/16] = -23

$A={\displaystyle \underset{\pi /4}{\overset{5\pi /4}{\int }}\left(sinx-cosx\right)dx={\left[-cosx-sinx\right]}_{\pi /4}^{5\pi /4}}=\left(\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{2}}\right)-\left(-\frac{1}{\sqrt[]{2}}-\frac{1}{\sqrt[]{2}}\right)$

A =    $2\sqrt[]{2}$

A⁴ = 64

sin 2qθ+ tan 2θ> 0

$\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{2tan\theta }{1-ta{n}^2\theta }>0$           

Let tan q = x

  $\frac{2x}{1+x^2}+\frac{2x}{1-x^2}>0$          

$tan\theta <-1or0<tan\theta <1$          

  $\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$          

 $cosec\left(2co{t}^{-1}\left(5\right)+co{s}^{-1}\left(\frac{4}{5}\right)\right)$

Let cot^-1 (5) = and cos^-1 $\left(\frac{4}{5}\right)=\alpha$

cot = 5 cos = $\frac{4}{5}$

$=cosec\left(2\theta +\alpha \right)=\frac{1}{sin\left(2\theta +\alpha \right)}$

$as\{\begin{array}{l}sin2\theta =\frac{2tan\theta }{1+ta{n}^2\theta }=\frac{2\left(\frac{1}{5}\right)}{1+\frac{1}{25}}=\frac{5}{13}\\ cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-\frac{1}{25}}{1+\frac{1}{25}}=\frac{12}{13}\end{array}$