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The prefix is actually "circ-" means "around."
The word "circ" originates from the Latin word "circus," meaning "circle" or "ring." This Latin term is derived from the Greek word "kirkos," which also means "circle." In English, "circ" is commonly used as a prefix in words related to circular motion or encirclement, such as "circumference" and "circumstance."
A regular decagon can rotate onto itself at angles that are multiples of ( \frac{360^\circ}{10} ), which is ( 36^\circ ). This means it can rotate by ( 0^\circ ), ( 36^\circ ), ( 72^\circ ), ( 108^\circ ), ( 144^\circ ), ( 180^\circ ), ( 216^\circ ), ( 252^\circ ), ( 288^\circ ), and ( 324^\circ ). In total, there are 10 distinct angles (including ( 0^\circ )) at which the decagon can map onto itself.
What does Ca. stand for? ca. is also abbreviated c., cir., or circ., and stands for circa which is from Latin and means "around," "approximately," "about." Sometimes it is italicized.
To find the value of (\tan(15^\circ) \tan(195^\circ)), we can use the identity (\tan(195^\circ) = \tan(15^\circ + 180^\circ) = \tan(15^\circ)). Thus, (\tan(195^\circ) = \tan(15^\circ)). Consequently, (\tan(15^\circ) \tan(195^\circ) = \tan(15^\circ) \tan(15^\circ) = \tan^2(15^\circ)). The exact value of (\tan^2(15^\circ)) can be computed, but it is important to note that it will yield a positive value.
To find the value of ( \cos^2 67^\circ - \sin^2 23^\circ ), we can use the identity ( \cos^2 \theta = 1 - \sin^2 \theta ). Since ( \sin 23^\circ = \cos 67^\circ ) (because ( 23^\circ + 67^\circ = 90^\circ )), we have ( \sin^2 23^\circ = \cos^2 67^\circ ). Thus, ( \cos^2 67^\circ - \sin^2 23^\circ = \cos^2 67^\circ - \cos^2 67^\circ = 0 ). Therefore, the value is ( 0 ).
The word actually comes from the Latin for "Mars," as in the god of war.
The exact value of (\sin 165^\circ) can be calculated using the sine subtraction formula. Since (165^\circ = 180^\circ - 15^\circ), we have: [ \sin 165^\circ = \sin(180^\circ - 15^\circ) = \sin 15^\circ ] The value of (\sin 15^\circ) can be derived from the formula (\sin(45^\circ - 30^\circ)), which gives: [ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} ] Thus, (\sin 165^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}).
The cofunction of cosine is sine. Therefore, the cofunction of (\cos 70^\circ) is (\sin(90^\circ - 70^\circ)), which simplifies to (\sin 20^\circ). Thus, (\cos 70^\circ = \sin 20^\circ).
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The cosine of 15 degrees can be calculated using the cosine subtraction formula: ( \cos(15^\circ) = \cos(45^\circ - 30^\circ) ). This gives us ( \cos(15^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ ). Plugging in the known values, ( \cos 45^\circ = \frac{\sqrt{2}}{2} ), ( \cos 30^\circ = \frac{\sqrt{3}}{2} ), ( \sin 45^\circ = \frac{\sqrt{2}}{2} ), and ( \sin 30^\circ = \frac{1}{2} ), we find that ( \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} ).
No, Latin comes from Italy.