Question

Prove that sin 54°+ cos72°
3/4​

Answer 1

Answer:

Cos(72) = Cos(90-18)=Sin(18) =

$\sqrt{5} - 1 \div 4$

----(a)

Sin(54) = Sin(90-36) = Cos(36) =

$\frac{ \sqrt{5 + 1} }{4}$

----(b)

a^2+b^2

(5+1+2root5/16)+(5+1-2root5/16)

=12/16=3/4

Answer 2

Cos(72) = Cos(90-18)=Sin(18) = \frac{ \sqrt{5} - 1 }{4}

4

5

−1

----(a)

Sin(54) = Sin(90-36) = Cos(36) = \frac{ \sqrt{5} + 1 }{4}

4

5

+1

----(b)

a^{2} + b^{2}a

2

+b

2

\frac{5 + 1 + 2 \sqrt{5} }{16}

16

5+1+2

5

+ \frac{5 + 1 - 2 \sqrt{5} }{16}

16

5+1−2

5

\frac{12}{16} = \frac{3}{4}

16

12

=

4

3

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