What is the value of sin^2(270)-cos^2(270)+tan^2(60) ?

The value of sin^2(270)-cos^2(270)+tan^2(60) is sin^2((3pi)/2)-cos^2((3pi)/2)+tan^2(pi/3)

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sin^2(270)-cos^2(270)+tan^2(60)

sin^{2} (27 0^{\circ}) - cos^{2} (27 0^{\circ}) + tan^{2} (6 0^{\circ})

sin^{2} (\frac{3 π}{2}) - cos^{2} (\frac{3 π}{2}) + tan^{2} (\frac{π}{3})

Decimal Notation

4

Solution steps

sin^{2} (27 0^{\circ}) - cos^{2} (27 0^{\circ}) + tan^{2} (6 0^{\circ})

Rewrite using trig identities:

- cos^{2} (27 0^{\circ}) + sin^{2} (27 0^{\circ}) = - cos (18 0^{\circ})

- cos^{2} (27 0^{\circ}) + sin^{2} (27 0^{\circ})

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

- cos^{2} (x) + sin^{2} (x) = - cos (2 x)

= - cos (2 \cdot 27 0^{\circ})

Simplify

= - cos (54 0^{\circ})

cos (54 0^{\circ}) = cos (18 0^{\circ})

cos (54 0^{\circ})

Rewrite

54 0^{\circ}

36 0^{\circ} + 18 0^{\circ}

= cos (36 0^{\circ} + 18 0^{\circ})

Apply the periodicity of

cos

cos (x + 36 0^{\circ}) = cos (x)

cos (36 0^{\circ} + 18 0^{\circ}) = cos (18 0^{\circ})

= cos (18 0^{\circ})

= - cos (18 0^{\circ})

= sin^{2} (27 0^{\circ}) - cos^{2} (27 0^{\circ}) + tan^{2} (6 0^{\circ})

8 cos (0)

tan (\frac{4}{5})

sin (\frac{\frac{π}{2}}{2})

- arctan (2)

5 sin (4 0^{\circ})

What is the value of sin^2(270)-cos^2(270)+tan^2(60) ?
The value of sin^2(270)-cos^2(270)+tan^2(60) is sin^2((3pi)/2)-cos^2((3pi)/2)+tan^2(pi/3)