find the quadratic equation whose roots are sine 18,cos 36.
Answers
Step-by-step explanation:
The quadratic equation is 4x^{2}-2\sqrt{5}x+4=0.4x
2
−2
5
x+4=0.
Step-by-step explanation:
We know that,
\sin 18=\dfrac{\sqrt{5}-1}{4}sin18=
4
5
−1
and
\cos 36=\dfrac{\sqrt{5}+1}{4}cos36=
4
5
+1
To find, the equation of whose roots are \sin 18sin18 and \cos 36cos36 =?
We know that,
The quadratic equation has
x^{2}-(\alpha+\beta)x+\alpha\beta=0x
2
−(α+β)x+αβ=0
∴ \sin 18+\cos 36=\dfrac{\sqrt{5}-1}{4}+\dfrac{\sqrt{5}+1}{4}sin18+cos36=
4
5
−1
+
4
5
+1
=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{4}=\dfrac{2\sqrt{5}}{4}=
4
5
−1+
5
+1
=
4
2
5
and,
\sin 18.\cos 36=\dfrac{\sqrt{5}-1}{4}.\dfrac{\sqrt{5}+1}{4}=\dfrac{\sqrt{5} ^{2}-1^{2}}{4}sin18.cos36=
4
5
−1
.
4
5
+1
=
4
5
2
−1
2
=\dfrac{5-1}{4}=1=
4
5−1
=1
The quadratic equation is:
x^{2}-(\dfrac{2\sqrt{5}}{4})x+1=0x
2
−(
4
2
5
)x+1=0
4x^{2}-2\sqrt{5}x+4=04x
2
−2
5
x+4=0
Hence, the quadratic equation is 4x^{2}-2\sqrt{5}x+4=0.4x
2
−2
5
x+4=0.