Question

if SinØ+Sin²Ø = 1, find the value of :-

Cos¹²Ø+3Cos¹⁰Ø+3Cos⁸Ø+Cos⁶Ø+2Cos⁴Ø+2Cos²Ø-2 ​

Answer 1

Answer:

★Question★

• If SinØ + Sin²Ø = 1, find the value of :-
• Cos¹²Ø + 3Cos¹⁰Ø +3 Cos⁸Ø + Cos⁶Ø + 2Cos⁴Ø +2 Cos²Ø - 2

★Answer★

★Given :-

• SinØ + Sin²Ø = 1

★To Find :-

• The value of Cos¹²Ø + 3Cos¹⁰Ø +3 Cos⁸Ø + Cos⁶Ø + 2Cos⁴Ø +2 Cos²Ø - 2

★Identity used

• sin²θ + cos²θ = 1
• (x + y)³ = x³ + y³ + 3xy (x + y)

★Solution:-

$\bf \:SinØ + Sin²Ø = 1$

$\bf\implies \:SinØ = 1 - Sin²Ø$

$\bf\implies \: SinØ = Cos²Ø ......(1)$

$\bf \:Consider \:</p><p>Cos¹²Ø + 3Cos¹⁰Ø +3 Cos⁸Ø + Cos⁶Ø + 2Cos⁴Ø +2 Cos²Ø - 2$

$\bf \:put \: Cos²Ø \: = SinØ \: in \: above \: we \: get$

$\bf\implies \: {(SinØ)}^{6} + 3 {(SinØ)}^{5} + 3 {(SinØ)}^{4} + {(SinØ)}^{3} + 2( {(SinØ)}^{2} + SinØ - 1)$

$\bf\implies \: {( {(SinØ)}^{2} )}^{3} + {(SinØ)}^{3} + 3 \times SinØ \times {((SinØ)}^{2} (SinØ + Sin²Ø) + 2(1 - 1)$

$\bf\implies \: {(SinØ + Sin²Ø)}^{3}$

$\bf\implies \: {(1)}^{3} = 1$

___________________________________________

Additional Information

Trigonometry Formula's

• sin θ = Opposite Side/Hypotenuse
• cos θ = Adjacent Side/Hypotenuse
• tan θ = Opposite Side/Adjacent Side
• sec θ = Hypotenuse/Adjacent Side
• cosec θ = Hypotenuse/Opposite Side
• cot θ = Adjacent Side/Opposite Side

◇ Reciprocal Identities

The Reciprocal Identities are given as:

• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• cot θ = 1/tan θ
• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ

◇ Co-function Identities

• sin (90°−x) = cos x
• cos (90°−x) = sin x
• tan (90°−x) = cot x
• cot (90°−x) = tan x
• sec (90°−x) = cosec x
• cosec (90°−x) = sec x

◇ Fundamental Trigonometric Identities

• sin²θ + cos²θ = 1
• sec²θ - tan²θ = 1
• cosec²θ - cot²θ = 1

Answer 2

Step-by-step explanation:

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