sinA+sin
2
A=1 and \displaystyle a \cos^{12} A + b \cos^8 A + c \cos^6 A - 1 = 0acos
12
A+bcos
8
A+ccos
6
A−1=0 then \displaystyle b+\frac{c}{a}+b = ?b+
a
c
+b=?
Answers
Answer:
sinA=1−sin
2
A
\displaystyle \sin A = \cos^2 AsinA=cos
2
A
\displaystyle \sin^2 A = \cos^4 Asin
2
A=cos
4
A
\displaystyle 1 - \cos ^2 A = \cos^4 A1−cos
2
A=cos
4
A
\displaystyle 1 = \cos^4 A + \cos^2 A1=cos
4
A+cos
2
A
\displaystyle 1^3 = (\cos^4 A + \cos^2 A)^31
3
=(cos
4
A+cos
2
A)
3
by formula \displaystyle (a+b)^3(a+b)
3
\displaystyle 1 = \cos^{12} A + 3 \cos^{10} A +3 \cos^8 A + \cos^6 A1=cos
12
A+3cos
10
A+3cos
8
A+cos
6
A
Transverse 1 onto the other side we get the condition as given as in the question. Comparing the variables we get \displaystyle a = 1a=1, \displaystyle b = 3b=3, \displaystyle c = 3c=3, \displaystyle d = 1d=1 hence the value of \displaystyle b +\frac{c}{a}+b = \frac{3 + 3}{1 + 1} = \frac{6}{2} = 3b+
a
c
+b=
1+1
3+3
=
2
6
=3:
Answer:
sinA=1−sin
2
A
\displaystyle \sin A = \cos^2 AsinA=cos
2
A
\displaystyle \sin^2 A = \cos^4 Asin
2
A=cos
4
A
\displaystyle 1 - \cos ^2 A = \cos^4 A1−cos
2
A=cos
4
A
\displaystyle 1 = \cos^4 A + \cos^2 A1=cos
4
A+cos
2
A
\displaystyle 1^3 = (\cos^4 A + \cos^2 A)^31
3
=(cos
4
A+cos
2
A)
3
by formula \displaystyle (a+b)^3(a+b)
3
\displaystyle 1 = \cos^{12} A + 3 \cos^{10} A +3 \cos^8 A + \cos^6 A1=cos
12
A+3cos
10
A+3cos
8
A+cos
6
A
Transverse 1 onto the other side we get the condition as given as in the question. Comparing the variables we get \displaystyle a = 1a=1, \displaystyle b = 3b=3, \displaystyle c = 3c=3, \displaystyle d = 1d=1 hence the value of \displaystyle b +\frac{c}{a}+b = \frac{3 + 3}{1 + 1} = \frac{6}{2} = 3b+
a
c
+b=
1+1
3+3
=
2
6
=3: