Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №2636219
А
Аноним
17 января 2025 04:22 · публичный вопрос
А
Аноним

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Ответ нейросети
17 января 2025 04:22

Sure, let's go through the problems step by step.

1. Compute:

a) \( \tan 45^\circ - \sin 30^\circ \)

  • \( \tan 45^\circ = 1 \)
  • \( \sin 30^\circ = 0.5 \)

So, \( 1 - 0.5 = 0.5 \).

b) \( \cos 60^\circ + \tan 45^\circ \)

  • \( \cos 60^\circ = 0.5 \)
  • \( \tan 45^\circ = 1 \)

So, \( 0.5 + 1 = 1.5 \).

2. Simplify the expressions:

a) \( (1 + \sin \alpha)(1 - \sin \alpha) \)

This is a difference of squares:

\[ 1 - \sin^2 \alpha = \cos^2 \alpha \]

b) \( \tan \alpha \cdot \cos \alpha \)

\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]

\[ \tan \alpha \cdot \cos \alpha = \sin \alpha \]

c) \( 1 + \cos^2 \alpha - \sin^2 \alpha \)

Using the identity \( \cos^2 \alpha = 1 - \sin^2 \alpha \):

\[ 1 + (1 - \sin^2 \alpha) - \sin^2 \alpha = 2 - 2\sin^2 \alpha = 2\cos^2 \alpha \]

3. Find the value:

a) \( \cos \alpha \), if \( \sin \alpha = 0.6 \)

Using \( \sin^2 \alpha + \cos^2 \alpha = 1 \):

\[ 0.6^2 + \cos^2 \alpha = 1 \] \[ 0.36 + \cos^2 \alpha = 1 \] \[ \cos^2 \alpha = 0.64 \] \[ \cos \alpha = \sqrt{0.64} = 0.8 \]

b) \( \sin \alpha \), if \( \cos \alpha = 0.8 \)

Similarly:

\[ \sin^2 \alpha + 0.8^2 = 1 \] \[ \sin^2 \alpha + 0.64 = 1 \] \[ \sin^2 \alpha = 0.36 \] \[ \sin \alpha = \sqrt{0.36} = 0.6 \]

I hope this helps! Let me know if you have further questions.