prove that: 2 cos 5 π 12 cos π 12 = 1 2 - transformation formulae
prove that: \[2\cos\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{1}{2}\]
\[lhs = 2\left( \cos \frac{5\pi}{12} \right) \left( \cos \frac{\pi}{12} \right)\]\[ = \cos \left( \frac{5\pi}{12} + \frac{\pi}{12} \right) + \cos \left( \frac{5\pi}{12} - \frac{\pi}{12} \right) \left[ \because 2 \cos a \cos b = \cos (a + b) + \cos (a - b) \right]\]\[ = \cos \frac{\pi}{2} + \cos \frac{\pi}{3}\]\[ = 0 + \frac{1}{2}\]\[ = \frac{1}{2}\]\[rhs = \frac{1}{2}\]hence, lhs = rhs
tan−1{2cos(2sin−121​)}=tan−1{2cos(2×6π​)}=tan−1{2cos(3π​)}=tan−1{2×21​}=tan−1{1}=4π​ the value of is we need, principal value of cos-1 x is [0,π]also, cos (2nπ + θ) = cos θ for all n є nso, doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) dear student,
1+ tan2x= 1+sin2x/cos2x
find common denominator which is cos2x
cos2x/cos2x +
sin2x/cos2x
(cos2x + sin2x)/cos2x
1+cot2x = 1+ cos2x/sin2x
common denominator is sin2x
sin2x/sin2x +
cos2x/sin2x
( sin2x + cos2x)/sin2x
1/(cosx2+ sin2x)/cos2x + 1/(
sin2x + cos2x)/sin2x
multiply the numerator and denominator by the reciprocal of the
bottom fractions to get
cos2x/(cosx2+ sin2x )+
sin2x/( cosx2+ sin2x)
since they now have a common denominator you can add the
tips
(cosx2+ sin2x)/(cosx2+
sin2x)
1/1 = 1

regards formula used/theory.there is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)we have,f(x) = 2 – xthen,zero of the polynomial f(x)means f(x) = 02 – x = 0-x = -2x = 2∴ zero of polynomial f(x) is 2conclusion.hence, the zero of the polynomial f(x) is 2 let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, 4x3 – 12x2 + 11x – 5 is divided by 2x – 1remainder theorem states that if p(x) is any polynomial and a is any real number and if p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).let p(x) = 4x3 – 12x2 + 11x – 5 and we have (2x – 1)the zero of (2x – 1) is now using remainder theorem,p(x) = 4x3 – 12x2 + 11x – 5 is divided by 2x – 1 then, is the remainderremainder = –2 The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved