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Giải thích các bước giải:

a.Ta có:

$\cos30^o+\cos60^o+\sin0^o$

$=\dfrac{\sqrt3}2+\dfrac12+0$

$=\dfrac{\sqrt3+1}2$

b.Ta có:

$\sin90^o+\cos90^o+\cos^20^o+\cot135^o$

$=1+0+1^2-1$

$=1$

c.Ta có:

$\cos^260^o+\tan^2120^o+\sin^260^o+\cot^260^o$

$=\cos^260^o+\tan^260^o+\sin^260^o+\cot^260^o$

$=(\cos^260^o+\sin^260^o)+(\tan^260^o+\cot^260^o)$

$=1+ (\sqrt3)^2+(\dfrac1{\sqrt3})^2$

$=\dfrac{13}3$

d.Ta có:

$\sin^215^o+\sin^235^o+\sin^255^o+\sin^275^o$

$=(\sin^215^o+\sin^275^o)+(\sin^235^o+\sin^255^o)$

$=(\sin^215^o+\cos^215^o)+(\sin^235^o+\cos^235^o)$

$=1+1$

$=2$